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def interpreter(code, iterations, width, height): grid = [[0 for r in range(width)] for c in range(height)] t = iterations (w, h) = (width, height) (stack, bracket_pos) = ([], {}) for (i, c) in enumerate(code): if c == '[': stack.append(i) elif c == ']': bracket_pos[i] = stack[-1] bracket_pos[stack.pop()] = i (a, b, p) = (0, 0, 0) while t > 0 and p < len(code): if code[p] == 'e': b += 1 elif code[p] == 'w': b -= 1 elif code[p] == 's': a += 1 elif code[p] == 'n': a -= 1 elif code[p] == '*': grid[a % h][b % w] ^= 1 elif code[p] == '[': if grid[a % h][b % w] == 0: p = bracket_pos[p] elif code[p] == ']': if grid[a % h][b % w] == 1: p = bracket_pos[p] else: t += 1 t -= 1 p += 1 return '\r\n'.join((''.join(map(str, g)) for g in grid))

python

# Esolang Interpreters #3 - Custom Paintfuck Interpreter ## About this Kata Series "Esolang Interpreters" is a Kata Series that originally began as three separate, independent esolang interpreter Kata authored by [@donaldsebleung](http://codewars.com/users/donaldsebleung) which all shared a similar format and were all somewhat inter-related. Under the influence of [a fellow Codewarrior](https://www.codewars.com/users/nickkwest), these three high-level inter-related Kata gradually evolved into what is known today as the "Esolang Interpreters" series. This series is a high-level Kata Series designed to challenge the minds of bright and daring programmers by implementing interpreters for various [esoteric programming languages/Esolangs](http://esolangs.org), mainly [Brainfuck](http://esolangs.org/wiki/Brainfuck) derivatives but not limited to them, given a certain specification for a certain Esolang. Perhaps the only exception to this rule is the very first Kata in this Series which is intended as an introduction/taster to the world of esoteric programming languages and writing interpreters for them. ## The Language Paintfuck is a [borderline-esoteric programming language/Esolang](http://esolangs.org) which is a derivative of [Smallfuck](http://esolangs.org/wiki/Smallfuck) (itself a derivative of the famous [Brainfuck](http://esolangs.org/wiki/Brainfuck)) that uses a two-dimensional data grid instead of a one-dimensional tape. Valid commands in Paintfuck include: - `n` - Move data pointer north (up) - `e` - Move data pointer east (right) - `s` - Move data pointer south (down) - `w` - Move data pointer west (left) - `*` - Flip the bit at the current cell (same as in Smallfuck) - `[` - Jump past matching `]` if bit under current pointer is `0` (same as in Smallfuck) - `]` - Jump back to the matching `[` (if bit under current pointer is nonzero) (same as in Smallfuck) The specification states that any non-command character (i.e. any character other than those mentioned above) should simply be ignored. The output of the interpreter is the two-dimensional data grid itself, best as animation as the interpreter is running, but at least a representation of the data grid itself after a certain number of iterations (explained later in task). In current implementations, the 2D datagrid is finite in size with toroidal (wrapping) behaviour. This is one of the few major differences of Paintfuck from Smallfuck as Smallfuck terminates (normally) whenever the pointer exceeds the bounds of the tape. Similar to Smallfuck, Paintfuck is Turing-complete **if and only if** the 2D data grid/canvas were unlimited in size. However, since the size of the data grid is defined to be finite, it acts like a finite state machine. More info on this Esolang can be found [here](http://esolangs.org/wiki/Paintfuck). ## The Task Your task is to implement a custom Paintfuck interpreter `interpreter()`/`Interpret` which accepts the following arguments in the specified order: 1. `code` - **Required**. The Paintfuck code to be executed, passed in as a string. May contain comments (non-command characters), in which case your interpreter should simply ignore them. If empty, simply return the initial state of the data grid. 2. `iterations` - **Required**. A non-negative integer specifying the number of iterations to be performed before the final state of the data grid is returned. See notes for definition of 1 iteration. If equal to zero, simply return the initial state of the data grid. 3. `width` - **Required**. The width of the data grid in terms of the number of data cells in each row, passed in as a positive integer. 4. `height` - **Required**. The height of the data grid in cells (i.e. number of rows) passed in as a positive integer. A few things to note: - Your interpreter should treat all command characters as **case-sensitive** so `N`, `E`, `S` and `W` are **not** valid command characters - Your interpreter should initialize all cells within the data grid to a value of `0` regardless of the width and height of the grid - In this implementation, your pointer must always start at the **top-left hand corner** of the data grid (i.e. first row, first column). This is important as some implementations have the data pointer starting at the middle of the grid. - One iteration is defined as one step in the program, i.e. the number of command characters evaluated. For example, given a program `nessewnnnewwwsswse` and an iteration count of `5`, your interpreter should evaluate `nesse` before returning the final state of the data grid. **Non-command characters should not count towards the number of iterations.** - Regarding iterations, the act of skipping to the matching `]` when a `[` is encountered (or vice versa) is considered to be **one** iteration regardless of the number of command characters in between. The next iteration then commences at the command **right after** the matching `]` (or `[`). - Your interpreter should terminate normally and return the final state of the 2D data grid whenever **any** of the mentioned conditions become true: (1) All commands have been considered left to right, or (2) Your interpreter has already performed the number of iterations specified in the second argument. - The return value of your interpreter should be a representation of the final state of the 2D data grid where each row is separated from the next by a CRLF (`\r\n`). For example, if the final state of your datagrid is ``` [ [1, 0, 0], [0, 1, 0], [0, 0, 1] ] ``` ... then your return string should be `"100\r\n010\r\n001"`. Good luck :D ## Kata in this Series 1. [Esolang Interpreters #1 - Introduction to Esolangs and My First Interpreter (MiniStringFuck)](https://www.codewars.com/kata/esolang-interpreters-number-1-introduction-to-esolangs-and-my-first-interpreter-ministringfuck) 2. [Esolang Interpreters #2 - Custom Smallfuck Interpreter](http://codewars.com/kata/esolang-interpreters-number-2-custom-smallfuck-interpreter) 3. **Esolang Interpreters #3 - Custom Paintfuck Interpreter** 4. [Esolang Interpreters #4 - Boolfuck Interpreter](http://codewars.com/kata/esolang-interpreters-number-4-boolfuck-interpreter)

taco

def interpreter(code, iterations, width, height): matrix = [[0 for i in range(width)] for i in range(height)] i = 0 iteration = 0 p = [0, 0] s = [] mate = {} for k in range(len(code)): c = code[k] if c == '[': s.append(k) if c == ']': m = s.pop() mate[m] = k mate[k] = m while iteration < iterations and i < len(code): c = code[i] if c == '*': matrix[p[1]][p[0]] ^= 1 elif c == 'n': p[1] = (p[1] - 1) % height elif c == 'e': p[0] = (p[0] + 1) % width elif c == 's': p[1] = (p[1] + 1) % height elif c == 'w': p[0] = (p[0] - 1) % width elif c == '[': if not matrix[p[1]][p[0]]: i = mate[i] elif c == ']': if matrix[p[1]][p[0]]: i = mate[i] else: iteration -= 1 i += 1 iteration += 1 return '\r\n'.join([''.join([str(matrix[y][x]) for x in range(width)]) for y in range(height)])

python

# Esolang Interpreters #3 - Custom Paintfuck Interpreter ## About this Kata Series "Esolang Interpreters" is a Kata Series that originally began as three separate, independent esolang interpreter Kata authored by [@donaldsebleung](http://codewars.com/users/donaldsebleung) which all shared a similar format and were all somewhat inter-related. Under the influence of [a fellow Codewarrior](https://www.codewars.com/users/nickkwest), these three high-level inter-related Kata gradually evolved into what is known today as the "Esolang Interpreters" series. This series is a high-level Kata Series designed to challenge the minds of bright and daring programmers by implementing interpreters for various [esoteric programming languages/Esolangs](http://esolangs.org), mainly [Brainfuck](http://esolangs.org/wiki/Brainfuck) derivatives but not limited to them, given a certain specification for a certain Esolang. Perhaps the only exception to this rule is the very first Kata in this Series which is intended as an introduction/taster to the world of esoteric programming languages and writing interpreters for them. ## The Language Paintfuck is a [borderline-esoteric programming language/Esolang](http://esolangs.org) which is a derivative of [Smallfuck](http://esolangs.org/wiki/Smallfuck) (itself a derivative of the famous [Brainfuck](http://esolangs.org/wiki/Brainfuck)) that uses a two-dimensional data grid instead of a one-dimensional tape. Valid commands in Paintfuck include: - `n` - Move data pointer north (up) - `e` - Move data pointer east (right) - `s` - Move data pointer south (down) - `w` - Move data pointer west (left) - `*` - Flip the bit at the current cell (same as in Smallfuck) - `[` - Jump past matching `]` if bit under current pointer is `0` (same as in Smallfuck) - `]` - Jump back to the matching `[` (if bit under current pointer is nonzero) (same as in Smallfuck) The specification states that any non-command character (i.e. any character other than those mentioned above) should simply be ignored. The output of the interpreter is the two-dimensional data grid itself, best as animation as the interpreter is running, but at least a representation of the data grid itself after a certain number of iterations (explained later in task). In current implementations, the 2D datagrid is finite in size with toroidal (wrapping) behaviour. This is one of the few major differences of Paintfuck from Smallfuck as Smallfuck terminates (normally) whenever the pointer exceeds the bounds of the tape. Similar to Smallfuck, Paintfuck is Turing-complete **if and only if** the 2D data grid/canvas were unlimited in size. However, since the size of the data grid is defined to be finite, it acts like a finite state machine. More info on this Esolang can be found [here](http://esolangs.org/wiki/Paintfuck). ## The Task Your task is to implement a custom Paintfuck interpreter `interpreter()`/`Interpret` which accepts the following arguments in the specified order: 1. `code` - **Required**. The Paintfuck code to be executed, passed in as a string. May contain comments (non-command characters), in which case your interpreter should simply ignore them. If empty, simply return the initial state of the data grid. 2. `iterations` - **Required**. A non-negative integer specifying the number of iterations to be performed before the final state of the data grid is returned. See notes for definition of 1 iteration. If equal to zero, simply return the initial state of the data grid. 3. `width` - **Required**. The width of the data grid in terms of the number of data cells in each row, passed in as a positive integer. 4. `height` - **Required**. The height of the data grid in cells (i.e. number of rows) passed in as a positive integer. A few things to note: - Your interpreter should treat all command characters as **case-sensitive** so `N`, `E`, `S` and `W` are **not** valid command characters - Your interpreter should initialize all cells within the data grid to a value of `0` regardless of the width and height of the grid - In this implementation, your pointer must always start at the **top-left hand corner** of the data grid (i.e. first row, first column). This is important as some implementations have the data pointer starting at the middle of the grid. - One iteration is defined as one step in the program, i.e. the number of command characters evaluated. For example, given a program `nessewnnnewwwsswse` and an iteration count of `5`, your interpreter should evaluate `nesse` before returning the final state of the data grid. **Non-command characters should not count towards the number of iterations.** - Regarding iterations, the act of skipping to the matching `]` when a `[` is encountered (or vice versa) is considered to be **one** iteration regardless of the number of command characters in between. The next iteration then commences at the command **right after** the matching `]` (or `[`). - Your interpreter should terminate normally and return the final state of the 2D data grid whenever **any** of the mentioned conditions become true: (1) All commands have been considered left to right, or (2) Your interpreter has already performed the number of iterations specified in the second argument. - The return value of your interpreter should be a representation of the final state of the 2D data grid where each row is separated from the next by a CRLF (`\r\n`). For example, if the final state of your datagrid is ``` [ [1, 0, 0], [0, 1, 0], [0, 0, 1] ] ``` ... then your return string should be `"100\r\n010\r\n001"`. Good luck :D ## Kata in this Series 1. [Esolang Interpreters #1 - Introduction to Esolangs and My First Interpreter (MiniStringFuck)](https://www.codewars.com/kata/esolang-interpreters-number-1-introduction-to-esolangs-and-my-first-interpreter-ministringfuck) 2. [Esolang Interpreters #2 - Custom Smallfuck Interpreter](http://codewars.com/kata/esolang-interpreters-number-2-custom-smallfuck-interpreter) 3. **Esolang Interpreters #3 - Custom Paintfuck Interpreter** 4. [Esolang Interpreters #4 - Boolfuck Interpreter](http://codewars.com/kata/esolang-interpreters-number-4-boolfuck-interpreter)

taco

n = int(input()) arr = list(map(int, input().split())) tracker = [[-1] * (n + 1) for _ in range(2024)] d = [[] for _ in range(n)] for (j, v) in enumerate(arr): tracker[v][j] = j d[j].append(j) for v in range(1, 2024): for i in range(n): j = tracker[v][i] h = tracker[v][j + 1] if j != -1 else -1 if j != -1 and h != -1: tracker[v + 1][i] = h d[i].append(h) a = [_ for _ in range(1, n + 1)] for s in range(n): for tracker in d[s]: a[tracker] = min(a[tracker], a[s - 1] + 1 if s > 0 else 1) print(a[n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys input = sys.stdin.readline n = int(input()) a = list(map(int, input().split())) INF = 10 ** 9 dp = [[INF] * (n + 1) for i in range(n + 1)] val = [[-1] * (n + 1) for i in range(n + 1)] for i in range(n): dp[i][i + 1] = 1 val[i][i + 1] = a[i] for l in range(2, n + 1): for i in range(n - l + 1): j = i + l for k in range(i + 1, j): if dp[i][k] == dp[k][j] == 1 and val[i][k] == val[k][j]: dp[i][j] = 1 val[i][j] = val[i][k] + 1 else: dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]) print(dp[0][n])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys readline = sys.stdin.buffer.readline N = int(readline()) A = list(map(int, readline().split())) dp = [[0] * N for _ in range(N)] for j in range(N): dp[j][0] = A[j] for l in range(1, N): for j in range(l, N): for k in range(j - l, j): if dp[k][k - j + l] == dp[j][j - k - 1] > 0: dp[j][l] = 1 + dp[j][j - k - 1] break dp = [None] + dp Dp = [0] * (N + 1) for j in range(1, N + 1): res = N for l in range(j): if dp[j][l]: res = min(res, 1 + Dp[j - l - 1]) Dp[j] = res print(Dp[N])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) A = list(map(int, input().split())) INF = 10 ** 3 dp = [[INF] * (n + 1) for _ in range(n + 1)] val = [[0] * (n + 1) for _ in range(n + 1)] for i in range(n): dp[i][i + 1] = 1 for i in range(n): val[i][i + 1] = A[i] for d in range(2, n + 1): for i in range(n + 1 - d): j = i + d for k in range(i + 1, j): if dp[i][k] == 1 and dp[k][j] == 1 and (val[i][k] == val[k][j]): dp[i][j] = min(dp[i][j], 1) val[i][j] = val[i][k] + 1 else: dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]) print(dp[0][n])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

from collections import defaultdict, deque from heapq import heappush, heappop from bisect import bisect_left, bisect_right import sys, itertools, math sys.setrecursionlimit(10 ** 5) input = sys.stdin.readline sqrt = math.sqrt def LI(): return list(map(int, input().split())) def LF(): return list(map(float, input().split())) def LI_(): return list(map(lambda x: int(x) - 1, input().split())) def II(): return int(input()) def IF(): return float(input()) def S(): return input().rstrip() def LS(): return S().split() def IR(n): res = [None] * n for i in range(n): res[i] = II() return res def LIR(n): res = [None] * n for i in range(n): res[i] = LI() return res def FR(n): res = [None] * n for i in range(n): res[i] = IF() return res def LIR(n): res = [None] * n for i in range(n): res[i] = IF() return res def LIR_(n): res = [None] * n for i in range(n): res[i] = LI_() return res def SR(n): res = [None] * n for i in range(n): res[i] = S() return res def LSR(n): res = [None] * n for i in range(n): res[i] = LS() return res mod = 1000000007 inf = float('INF') def solve(): n = II() a = LI() dp = [[None for i in range(n + 1)] for i in range(n + 1)] for i in range(n): dp[i][i + 1] = [a[i], a[i], 1] dp[i + 1][i] = [a[i], a[i], 1] for i in range(2, n + 1): for l in range(n - i + 1): tmp = [-inf, inf, inf] r = l + i dpl = dp[l] dpr = dp[r] for m in range(l + 1, r): lm = dpl[m] mr = dpr[m] lr = lm[2] + mr[2] - (lm[1] == mr[0]) if lr < tmp[2]: tmp[2] = lr if lm[1] == mr[0]: if lm[2] == 1: tmp[0] = lm[0] + 1 else: tmp[0] = lm[0] if mr[2] == 1: tmp[1] = mr[1] + 1 else: tmp[1] = mr[1] else: tmp[0] = lm[0] tmp[1] = mr[1] dp[l][r] = tmp dp[r][l] = tmp print(dp[0][n][2]) return solve()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import io import os import sys from functools import lru_cache from collections import defaultdict sys.setrecursionlimit(10 ** 5) def solve(N, A): valToLeftRight = defaultdict(lambda : defaultdict(set)) valToRightLeft = defaultdict(lambda : defaultdict(set)) for (i, x) in enumerate(A): valToLeftRight[x][i].add(i) valToRightLeft[x][i].add(i) maxVal = 1000 + 10 for val in range(maxVal): for (l, rights) in valToLeftRight[val - 1].items(): for r in rights: l2 = r + 1 if l2 in valToLeftRight[val - 1]: for r2 in valToLeftRight[val - 1][l2]: assert l <= r assert r + 1 == l2 assert l2 <= r2 valToLeftRight[val][l].add(r2) valToRightLeft[val][r2].add(l) r2 = l - 1 if r2 in valToRightLeft[val - 1]: for l2 in valToRightLeft[val - 1][r2]: assert l2 <= r2 assert r2 == l - 1 assert l <= r valToLeftRight[val][l2].add(r) valToRightLeft[val][r].add(l2) intervals = defaultdict(list) for val in range(maxVal): for (l, rights) in valToLeftRight[val].items(): for r in rights: intervals[l].append(r) @lru_cache(maxsize=None) def getBest(left): if left == N: return 0 best = float('inf') for right in intervals[left]: best = min(best, 1 + getBest(right + 1)) return best return getBest(0) input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline (N,) = list(map(int, input().split())) A = list(map(int, input().split())) ans = solve(N, A) print(ans)

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys input = sys.stdin.readline n = int(input()) A = list(map(int, input().split())) DP = [[-1] * (n + 1) for i in range(n + 1)] for i in range(n): DP[i][i] = A[i] for mid in range(1, n): for i in range(n): j = i + mid if j == n: break for k in range(i, j + 1): if DP[i][k] == DP[k + 1][j] and DP[i][k] != -1: DP[i][j] = DP[i][k] + 1 ANS = [2000] * (n + 1) ANS.append(0) for i in range(n): ANS[i] = min(ANS[i], ANS[i - 1] + 1) for j in range(i, n): if DP[i][j] != -1: ANS[j] = min(ANS[j], ANS[i - 1] + 1) print(ANS[n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) b = [int(_) for _ in input().split()] e = [[-1] * (n + 1) for _ in range(2024)] d = [[] for _ in range(n)] for (j, v) in enumerate(b): e[v][j] = j d[j].append(j) for v in range(1, 2024): for i in range(n): j = e[v][i] h = e[v][j + 1] if j != -1 else -1 if j != -1 and h != -1: e[v + 1][i] = h d[i].append(h) a = [_ for _ in range(1, n + 1)] for s in range(n): for e in d[s]: a[e] = min(a[e], a[s - 1] + 1 if s > 0 else 1) print(a[n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) a = list(map(int, input().split())) dp = [[505] * n for _ in range(n)] Max = [[0] * n for _ in range(n)] for i in range(n): dp[i][i] = 1 Max[i][i] = a[i] for len in range(1, n + 1): for i in range(n - len + 1): j = i + len - 1 for k in range(i, j): dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]) if dp[i][k] == 1 and dp[k + 1][j] == 1 and (Max[i][k] == Max[k + 1][j]): dp[i][j] = 1 Max[i][j] = Max[i][k] + 1 print(dp[0][n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys, math import io, os from bisect import bisect_left as bl, bisect_right as br, insort from collections import defaultdict as dd, deque, Counter def data(): return sys.stdin.readline().strip() def mdata(): return list(map(int, data().split())) def outl(var): sys.stdout.write(' '.join(map(str, var)) + '\n') def out(var): sys.stdout.write(str(var) + '\n') from decimal import Decimal INF = float('inf') mod = int(1000000000.0) + 7 def cal(l, r): if l == r: dp1[l][r] = 1 dp2[l][r] = a[l] if dp1[l][r]: return dp1[l][r] for i in range(l, r): if cal(l, i) == 1 and cal(i + 1, r) == 1 and (dp2[l][i] == dp2[i + 1][r]): dp1[l][r] = 1 dp2[l][r] = dp2[l][i] + 1 if not dp2[l][r]: dp1[l][r] = 2 return dp1[l][r] def cal2(l, r): if dp1[l][r] == 1: dp3[l][r] = 1 return 1 elif dp3[l][r]: return dp3[l][r] ans = INF for i in range(l, r): ans = min(cal2(l, i) + cal2(i + 1, r), ans) dp3[l][r] = ans return ans n = int(data()) a = mdata() ans = [n] dp1 = [[0] * n for i in range(n)] dp2 = [[0] * n for i in range(n)] dp3 = [[0] * n for i in range(n)] cal(0, n - 1) cal2(0, n - 1) out(dp3[0][n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) b = list(map(int, input().split(' '))) e = [[-1] * (n + 1) for _ in range(2048)] d = [[] for _ in range(n)] for (i, v) in enumerate(b): e[v][i] = i d[i].append(i) for v in range(1, 2048): for i in range(n): j = e[v][i] if j != -1: h = e[v][j + 1] else: h = -1 if j != -1 and h != -1: e[v + 1][i] = h d[i].append(h) a = [_ for _ in range(1, n + 1)] for s in range(n): for e in d[s]: if s > 0: temp = a[s - 1] + 1 else: temp = 1 a[e] = min(a[e], temp) print(a[n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import os import sys from io import BytesIO, IOBase BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = 'x' in file.mode or 'r' not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b'\n') + (not b) ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) (self.buffer.truncate(0), self.buffer.seek(0)) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode('ascii')) self.read = lambda : self.buffer.read().decode('ascii') self.readline = lambda : self.buffer.readline().decode('ascii') (sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout)) input = lambda : sys.stdin.readline().rstrip('\r\n') from math import gcd, ceil def prod(a, mod=10 ** 9 + 7): ans = 1 for each in a: ans = ans * each % mod return ans def lcm(a, b): return a * b // gcd(a, b) def binary(x, length=16): y = bin(x)[2:] return y if len(y) >= length else '0' * (length - len(y)) + y for _ in range(int(input()) if not True else 1): n = int(input()) a = list(map(int, input().split())) dp = [[False] * (n + 2) for i in range(n + 2)] dp2 = [[600] * (n + 2) for i in range(n + 2)] for i in range(n): dp[i][i] = a[i] dp2[i][i] = 1 for diff in range(1, n): for i in range(n - diff): for j in range(i, i + diff): if dp[i][j] == dp[j + 1][i + diff] and dp[i][j]: dp[i][i + diff] = dp[i][j] + 1 dp2[i][i + diff] = 1 dp2[i][i + diff] = min(dp2[i][i + diff], dp2[i][j] + dp2[j + 1][i + diff]) if not dp2[i][i + diff]: dp2[i][i + diff] = min(dp2[i + 1][i + diff] + 1, dp2[i][i + diff - 1] + 1) print(dp2[0][n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

N = int(input()) X = list(map(int, input().split())) from collections import defaultdict dp1 = defaultdict(lambda : -1) M = 1001 def ec(i, j): return i * M + j for i in range(N): dp1[ec(i, i + 1)] = X[i] for i in range(2, N + 1): for j in range(N - i + 1): for k in range(1, i): (u, v) = (dp1[ec(j, j + k)], dp1[ec(j + k, j + i)]) if u != -1 and v != -1 and (u == v): dp1[ec(j, j + i)] = u + 1 break dp2 = [0] * (N + 1) for i in range(N): dp2[i + 1] = dp2[i] + 1 for j in range(i + 1): if dp1[ec(j, i + 1)] == -1: continue dp2[i + 1] = min(dp2[i + 1], dp2[j] + 1) print(dp2[-1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys input = lambda : sys.stdin.readline().rstrip() import copy n = int(input()) A = [int(i) for i in input().split()] inf = float('inf') DP = [[inf] * (n + 1) for _ in range(n + 1)] for j in range(1, n + 1): for i in range(n): if i + j > n: continue elif j == 1: DP[i][i + 1] = A[i] else: for k in range(i + 1, i + j): if DP[i][k] < 10000 and DP[k][i + j] < 10000: if DP[i][k] == DP[k][i + j]: DP[i][i + j] = DP[i][k] + 1 else: DP[i][i + j] = 20000 elif DP[i][k] < 10000: DP[i][i + j] = min(DP[i][i + j], 10000 + DP[k][i + j]) elif DP[k][i + j] < 10000: DP[i][i + j] = min(DP[i][i + j], DP[i][k] + 10000) else: DP[i][i + j] = min(DP[i][i + j], DP[i][k] + DP[k][i + j]) print(DP[0][n] // 10000 if DP[0][n] >= 10000 else 1)

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

N = int(input()) X = list(map(int, input().split())) from collections import defaultdict dp = defaultdict(lambda : -1) M = 1000001 for i in range(N): dp[i + M] = X[i] for i in range(2, N + 1): for j in range(N - i + 1): for k in range(1, i): (u, v) = (dp[j + M * k], dp[j + k + M * (i - k)]) if u == -1 or v == -1 or u != v: continue dp[j + M * i] = u + 1 break dp2 = [0] * (N + 1) for i in range(N): dp2[i + 1] = dp2[i] + 1 for j in range(i + 1): if dp[j + (i + 1 - j) * M] == -1: continue dp2[i + 1] = min(dp2[i + 1], dp2[j] + 1) print(dp2[-1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

N = int(input()) L = list(map(int, input().split())) DP = [[-1] * N for i in range(N)] for d in range(N): for s in range(N - d): e = s + d if s == e: DP[s][e] = L[s] continue for m in range(s, e): l = DP[s][m] r = DP[m + 1][e] if l == r and l != -1: DP[s][e] = max(DP[s][e], l + 1) DP2 = [i + 1 for i in range(N)] for i in range(N): if DP[0][i] != -1: DP2[i] = 1 continue for j in range(i): if DP[j + 1][i] != -1: DP2[i] = min(DP2[i], DP2[j] + 1) print(DP2[N - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

printn = lambda x: print(x, end='') inn = lambda : int(input()) inl = lambda : list(map(int, input().split())) inm = lambda : map(int, input().split()) ins = lambda : input().strip() DBG = True and False BIG = 10 ** 18 R = 10 ** 9 + 7 def ddprint(x): if DBG: print(x) def dp(l, r): if dpa[l][r] >= 0: return dpa[l][r] elif l == r: dpa[l][r] = a[l] else: dpa[l][r] = 0 for j in range(l, r): x = dp(l, j) y = dp(j + 1, r) if 0 < x == y: dpa[l][r] = x + 1 return dpa[l][r] n = inn() a = inl() dpa = [[-1] * (n + 1) for i in range(n + 1)] dp2 = [BIG] * (n + 1) dp2[0] = 1 for i in range(n): if dp(0, i) > 0: dp2[i] = 1 else: mn = i + 1 for j in range(i): x = dp(j + 1, i) if 0 < x: mn = min(mn, dp2[j] + 1) dp2[i] = mn for i in range(n): ddprint(dpa[i]) ddprint(dp2) print(dp2[n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys input = sys.stdin.readline n = int(input()) a = list(map(int, input().split())) dp = [[1000] * (n + 1) for i in range(n + 1)] val = [[0] * (n + 1) for i in range(n + 1)] for i in range(n): dp[i][i + 1] = 1 val[i][i + 1] = a[i] for p in range(2, n + 1): for i in range(n - p + 1): j = i + p for k in range(i + 1, j): if dp[i][k] == dp[k][j] == 1 and val[i][k] == val[k][j]: dp[i][j] = 1 val[i][j] = val[i][k] + 1 else: dp[i][j] = min(dp[i][j], dp[i][k] + dp[k][j]) print(dp[0][n])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

rr = lambda : input().rstrip() rri = lambda : int(rr()) rrm = lambda : list(map(int, rr().split())) from functools import lru_cache memo = lru_cache(None) from sys import setrecursionlimit as srl srl(10 ** 5) def solve(N, A): @memo def dp(i, j, left=0): if i == j: if left == 0: return 1 if A[i] == left: return 1 return 2 if i > j: return 0 if left == 0 else 1 ans = 1 + dp(i + 1, j, A[i]) if left >= 1: stack = [] for k in range(i, j + 1): stack.append(A[k]) while len(stack) >= 2 and stack[-1] == stack[-2]: stack.pop() stack[-1] += 1 if len(stack) == 1 and left == stack[-1]: cand = dp(k + 1, j, left + 1) if cand < ans: ans = cand return ans return dp(1, N - 1, A[0]) print(solve(rri(), rrm()))

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) a = list(map(int, input().split())) grip = [[-1] * (n - i) for i in range(n)] grip[0] = a.copy() for level in range(1, n): for left in range(n - level): for split in range(level): pl = grip[level - split - 1][left] pr = grip[split][left + level - split] if pl == pr != -1: grip[level][left] = pl + 1 pref = [0] * (n + 1) for p in range(1, n + 1): x = n for j in range(p): l = pref[j] r = grip[p - j - 1][j] if r == -1: r = p - j else: r = 1 x = min(x, l + r) pref[p] = x print(pref[-1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import os, sys from io import BytesIO, IOBase from math import inf def main(): n = int(input()) a = [0] + list(map(int, input().split())) dp = [[[-1, 0] for _ in range(n + 1)] for _ in range(n + 1)] for i in range(1, n + 1): (dp[i][i][0], dp[i][i][1]) = (a[i], 1) for i in range(n - 1, 0, -1): for j in range(i + 1, n + 1): for k in range(j - i): (a, b) = (dp[i][i + k], dp[i + k + 1][j]) if a[1] and b[1] and (a[0] == b[0]): (dp[i][j][0], dp[i][j][1]) = (a[0] + 1, 1) break val = [0, 0] + [inf] * n for i in range(1, n + 1): for j in range(1, n + 1): if dp[i][j][1]: val[j + 1] = min(val[j + 1], val[i] + 1) print(val[-1]) BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = 'x' in file.mode or 'r' not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b'\n') + (not b) ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) (self.buffer.truncate(0), self.buffer.seek(0)) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode('ascii')) self.read = lambda : self.buffer.read().decode('ascii') self.readline = lambda : self.buffer.readline().decode('ascii') (sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout)) input = lambda : sys.stdin.readline().rstrip('\r\n') main()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys dp = [] a = [] def calcdp(l, r): global dp, a if l + 1 == r: dp[l][r] = a[l] return dp[l][r] if dp[l][r] != 0: return dp[l][r] dp[l][r] = -1 for k in range(l + 1, r): la = calcdp(l, k) ra = calcdp(k, r) if la > 0 and la == ra: dp[l][r] = la + 1 return dp[l][r] def solve(n): dp2 = [float('inf')] * (n + 1) dp2[0] = 0 for i in range(n): for j in range(i + 1, n + 1): if calcdp(i, j) > 0: dp2[j] = min(dp2[j], dp2[i] + 1) return dp2[n] def ip(): global dp, a n = int(sys.stdin.readline()) a = list(map(int, sys.stdin.readline().split())) a.append(0) dp = [] ll = [0] * (n + 1) for _ in range(n + 1): dp.append(list(ll)) print(solve(n)) ip()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

N = int(input()) arr = list(map(int, input().split())) dp = [[-1 for x in range(N)] for y in range(N)] for size in range(1, N + 1): for i in range(N - size + 1): j = i + size - 1 if i == j: dp[i][j] = arr[i] else: for k in range(i, j): if dp[i][k] != -1 and dp[i][k] == dp[k + 1][j]: dp[i][j] = dp[i][k] + 1 dp2 = [x + 1 for x in range(N)] for i in range(N): for k in range(i + 1): if dp[k][i] != -1: if k == 0: dp2[i] = 1 else: dp2[i] = min(dp2[i], dp2[k - 1] + 1) print(dp2[N - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) a = list(map(int, input().split())) dp = [[False] * (n + 1) for i in range(n + 1)] def solve(l, r): if dp[l][r]: return dp[l][r] if r - l == 1: dp[l][r] = (a[l], 1) return dp[l][r] tmp = 10 ** 9 for i in range(l + 1, r): if solve(l, i)[0] == -1 or solve(i, r)[0] == -1: tmp = min(tmp, dp[l][i][1] + dp[i][r][1]) elif solve(l, i) == solve(i, r): tmp = solve(l, i)[0] + 1 dp[l][r] = (tmp, 1) return dp[l][r] else: tmp = min(tmp, 2) dp[l][r] = (-1, tmp) return dp[l][r] solve(0, n) print(dp[0][n][1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys input = sys.stdin.readline N = int(input()) A = list(map(int, input().split())) dp = [[-1] * (N + 1) for _ in range(N + 1)] for l in range(N): dp[l][l + 1] = A[l] for d in range(2, N + 1): for l in range(N - d + 1): for t in range(1, d): if dp[l][l + t] == dp[l + t][l + d] and dp[l][l + t] != -1: dp[l][l + d] = dp[l][l + t] + 1 break dp2 = [i for i in range(N + 1)] for r in range(1, N + 1): if dp[0][r] != -1: dp2[r] = 1 for l in range(N): for r in range(l + 2, N + 1): if dp[l + 1][r] != -1: dp2[r] = min(dp2[l + 1] + 1, dp2[r]) else: dp2[r] = min(dp2[l + 1] + (r - l - 1), dp2[r]) print(dp2[N])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

from collections import Counter from collections import defaultdict import math import random import heapq as hq from math import sqrt import sys from functools import reduce def input(): return sys.stdin.readline().strip() def iinput(): return int(input()) def tinput(): return input().split() def rinput(): return map(int, tinput()) def rlinput(): return list(rinput()) mod = int(1000000000.0) + 7 def factors(n): return set(reduce(list.__add__, ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0))) n = iinput() a = rlinput() dp = [[0 for i in range(n)] for j in range(n)] for i in range(n): dp[i][i] = a[i] for l in range(n - 2, -1, -1): for r in range(l + 1, n): for k in range(l, r): if dp[l][k] == dp[k + 1][r] and dp[l][k] != 0: dp[l][r] = dp[l][k] + 1 squeeze = [float('inf')] * (n + 1) squeeze[0] = 0 for i in range(1, n + 1): for j in range(i): if dp[j][i - 1] != 0: squeeze[i] = min(squeeze[i], squeeze[j] + 1) print(squeeze[n])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

def divisors(M): d = [] i = 1 while M >= i ** 2: if M % i == 0: d.append(i) if i ** 2 != M: d.append(M // i) i = i + 1 return d def popcount(x): x = x - (x >> 1 & 1431655765) x = (x & 858993459) + (x >> 2 & 858993459) x = x + (x >> 4) & 252645135 x = x + (x >> 8) x = x + (x >> 16) return x & 127 def eratosthenes(n): res = [0 for i in range(n + 1)] prime = set([]) for i in range(2, n + 1): if not res[i]: prime.add(i) for j in range(1, n // i + 1): res[i * j] = 1 return prime def factorization(n): res = [] for p in prime: if n % p == 0: while n % p == 0: n //= p res.append(p) if n != 1: res.append(n) return res def euler_phi(n): res = n for x in range(2, n + 1): if x ** 2 > n: break if n % x == 0: res = res // x * (x - 1) while n % x == 0: n //= x if n != 1: res = res // n * (n - 1) return res def ind(b, n): res = 0 while n % b == 0: res += 1 n //= b return res def isPrimeMR(n): if n == 1: return 0 d = n - 1 d = d // (d & -d) L = [2, 3, 5, 7, 11, 13, 17] for a in L: t = d y = pow(a, t, n) if y == 1: continue while y != n - 1: y = y * y % n if y == 1 or t == n - 1: return 0 t <<= 1 return 1 def findFactorRho(n): from math import gcd m = 1 << n.bit_length() // 8 for c in range(1, 99): f = lambda x: (x * x + c) % n (y, r, q, g) = (2, 1, 1, 1) while g == 1: x = y for i in range(r): y = f(y) k = 0 while k < r and g == 1: ys = y for i in range(min(m, r - k)): y = f(y) q = q * abs(x - y) % n g = gcd(q, n) k += m r <<= 1 if g == n: g = 1 while g == 1: ys = f(ys) g = gcd(abs(x - ys), n) if g < n: if isPrimeMR(g): return g elif isPrimeMR(n // g): return n // g return findFactorRho(g) def primeFactor(n): i = 2 ret = {} rhoFlg = 0 while i * i <= n: k = 0 while n % i == 0: n //= i k += 1 if k: ret[i] = k i += 1 + i % 2 if i == 101 and n >= 2 ** 20: while n > 1: if isPrimeMR(n): (ret[n], n) = (1, 1) else: rhoFlg = 1 j = findFactorRho(n) k = 0 while n % j == 0: n //= j k += 1 ret[j] = k if n > 1: ret[n] = 1 if rhoFlg: ret = {x: ret[x] for x in sorted(ret)} return ret def divisors(n): res = [1] prime = primeFactor(n) for p in prime: newres = [] for d in res: for j in range(prime[p] + 1): newres.append(d * p ** j) res = newres res.sort() return res def xorfactorial(num): if num == 0: return 0 elif num == 1: return 1 elif num == 2: return 3 elif num == 3: return 0 else: x = baseorder(num) return 2 ** x * ((num - 2 ** x + 1) % 2) + function(num - 2 ** x) def xorconv(n, X, Y): if n == 0: res = [X[0] * Y[0] % mod] return res x = [digit[i] + X[i + 2 ** (n - 1)] for i in range(2 ** (n - 1))] y = [Y[i] + Y[i + 2 ** (n - 1)] for i in range(2 ** (n - 1))] z = [digit[i] - X[i + 2 ** (n - 1)] for i in range(2 ** (n - 1))] w = [Y[i] - Y[i + 2 ** (n - 1)] for i in range(2 ** (n - 1))] res1 = xorconv(n - 1, x, y) res2 = xorconv(n - 1, z, w) former = [(res1[i] + res2[i]) * inv for i in range(2 ** (n - 1))] latter = [(res1[i] - res2[i]) * inv for i in range(2 ** (n - 1))] former = list(map(lambda x: x % mod, former)) latter = list(map(lambda x: x % mod, latter)) return former + latter def merge_sort(A, B): (pos_A, pos_B) = (0, 0) (n, m) = (len(A), len(B)) res = [] while pos_A < n and pos_B < m: (a, b) = (A[pos_A], B[pos_B]) if a < b: res.append(a) pos_A += 1 else: res.append(b) pos_B += 1 res += A[pos_A:] res += B[pos_B:] return res class UnionFindVerSize: def __init__(self, N): self._parent = [n for n in range(0, N)] self._size = [1] * N self.group = N def find_root(self, x): if self._parent[x] == x: return x self._parent[x] = self.find_root(self._parent[x]) stack = [x] while self._parent[stack[-1]] != stack[-1]: stack.append(self._parent[stack[-1]]) for v in stack: self._parent[v] = stack[-1] return self._parent[x] def unite(self, x, y): gx = self.find_root(x) gy = self.find_root(y) if gx == gy: return self.group -= 1 if self._size[gx] < self._size[gy]: self._parent[gx] = gy self._size[gy] += self._size[gx] else: self._parent[gy] = gx self._size[gx] += self._size[gy] def get_size(self, x): return self._size[self.find_root(x)] def is_same_group(self, x, y): return self.find_root(x) == self.find_root(y) class WeightedUnionFind: def __init__(self, N): self.parent = [i for i in range(N)] self.size = [1 for i in range(N)] self.val = [0 for i in range(N)] self.flag = True self.edge = [[] for i in range(N)] def dfs(self, v, pv): stack = [(v, pv)] new_parent = self.parent[pv] while stack: (v, pv) = stack.pop() self.parent[v] = new_parent for (nv, w) in self.edge[v]: if nv != pv: self.val[nv] = self.val[v] + w stack.append((nv, v)) def unite(self, x, y, w): if not self.flag: return if self.parent[x] == self.parent[y]: self.flag = self.val[x] - self.val[y] == w return if self.size[self.parent[x]] > self.size[self.parent[y]]: self.edge[x].append((y, -w)) self.edge[y].append((x, w)) self.size[x] += self.size[y] self.val[y] = self.val[x] - w self.dfs(y, x) else: self.edge[x].append((y, -w)) self.edge[y].append((x, w)) self.size[y] += self.size[x] self.val[x] = self.val[y] + w self.dfs(x, y) class Dijkstra: class Edge: def __init__(self, _to, _cost): self.to = _to self.cost = _cost def __init__(self, V): self.G = [[] for i in range(V)] self._E = 0 self._V = V @property def E(self): return self._E @property def V(self): return self._V def add_edge(self, _from, _to, _cost): self.G[_from].append(self.Edge(_to, _cost)) self._E += 1 def shortest_path(self, s): import heapq que = [] d = [10 ** 15] * self.V d[s] = 0 heapq.heappush(que, (0, s)) while len(que) != 0: (cost, v) = heapq.heappop(que) if d[v] < cost: continue for i in range(len(self.G[v])): e = self.G[v][i] if d[e.to] > d[v] + e.cost: d[e.to] = d[v] + e.cost heapq.heappush(que, (d[e.to], e.to)) return d def Z_algorithm(s): N = len(s) Z_alg = [0] * N Z_alg[0] = N i = 1 j = 0 while i < N: while i + j < N and s[j] == s[i + j]: j += 1 Z_alg[i] = j if j == 0: i += 1 continue k = 1 while i + k < N and k + Z_alg[k] < j: Z_alg[i + k] = Z_alg[k] k += 1 i += k j -= k return Z_alg class BIT: def __init__(self, n, mod=0): self.BIT = [0] * (n + 1) self.num = n self.mod = mod def query(self, idx): res_sum = 0 mod = self.mod while idx > 0: res_sum += self.BIT[idx] if mod: res_sum %= mod idx -= idx & -idx return res_sum def update(self, idx, x): mod = self.mod while idx <= self.num: self.BIT[idx] += x if mod: self.BIT[idx] %= mod idx += idx & -idx return class dancinglink: def __init__(self, n, debug=False): self.n = n self.debug = debug self._left = [i - 1 for i in range(n)] self._right = [i + 1 for i in range(n)] self.exist = [True for i in range(n)] def pop(self, k): if self.debug: assert self.exist[k] L = self._left[k] R = self._right[k] if L != -1: if R != self.n: (self._right[L], self._left[R]) = (R, L) else: self._right[L] = self.n elif R != self.n: self._left[R] = -1 self.exist[k] = False def left(self, idx, k=1): if self.debug: assert self.exist[idx] res = idx while k: res = self._left[res] if res == -1: break k -= 1 return res def right(self, idx, k=1): if self.debug: assert self.exist[idx] res = idx while k: res = self._right[res] if res == self.n: break k -= 1 return res class SparseTable: def __init__(self, A, merge_func, ide_ele): N = len(A) n = N.bit_length() self.table = [[ide_ele for i in range(n)] for i in range(N)] self.merge_func = merge_func for i in range(N): self.table[i][0] = A[i] for j in range(1, n): for i in range(0, N - 2 ** j + 1): f = self.table[i][j - 1] s = self.table[i + 2 ** (j - 1)][j - 1] self.table[i][j] = self.merge_func(f, s) def query(self, s, t): b = t - s + 1 m = b.bit_length() - 1 return self.merge_func(self.table[s][m], self.table[t - 2 ** m + 1][m]) class BinaryTrie: class node: def __init__(self, val): self.left = None self.right = None self.max = val def __init__(self): self.root = self.node(-10 ** 15) def append(self, key, val): pos = self.root for i in range(29, -1, -1): pos.max = max(pos.max, val) if key >> i & 1: if pos.right is None: pos.right = self.node(val) pos = pos.right else: pos = pos.right elif pos.left is None: pos.left = self.node(val) pos = pos.left else: pos = pos.left pos.max = max(pos.max, val) def search(self, M, xor): res = -10 ** 15 pos = self.root for i in range(29, -1, -1): if pos is None: break if M >> i & 1: if xor >> i & 1: if pos.right: res = max(res, pos.right.max) pos = pos.left else: if pos.left: res = max(res, pos.left.max) pos = pos.right elif xor >> i & 1: pos = pos.right else: pos = pos.left if pos: res = max(res, pos.max) return res def solveequation(edge, ans, n, m): x = [0] * m used = [False] * n for v in range(n): if used[v]: continue y = dfs(v) if y != 0: return False return x def dfs(v): used[v] = True r = ans[v] for (to, dire, id) in edge[v]: if used[to]: continue y = dfs(to) if dire == -1: x[id] = y else: x[id] = -y r += y return r class Matrix: mod = 10 ** 9 + 7 def set_mod(m): Matrix.mod = m def __init__(self, L): self.row = len(L) self.column = len(L[0]) self._matrix = L for i in range(self.row): for j in range(self.column): self._matridigit[i][j] %= Matrix.mod def __getitem__(self, item): if type(item) == int: raise IndexError('you must specific row and column') elif len(item) != 2: raise IndexError('you must specific row and column') (i, j) = item return self._matridigit[i][j] def __setitem__(self, item, val): if type(item) == int: raise IndexError('you must specific row and column') elif len(item) != 2: raise IndexError('you must specific row and column') (i, j) = item self._matridigit[i][j] = val def __add__(self, other): if (self.row, self.column) != (other.row, other.column): raise SizeError('sizes of matrixes are different') res = [[0 for j in range(self.column)] for i in range(self.row)] for i in range(self.row): for j in range(self.column): res[i][j] = self._matridigit[i][j] + other._matridigit[i][j] res[i][j] %= Matrix.mod return Matrix(res) def __sub__(self, other): if (self.row, self.column) != (other.row, other.column): raise SizeError('sizes of matrixes are different') res = [[0 for j in range(self.column)] for i in range(self.row)] for i in range(self.row): for j in range(self.column): res[i][j] = self._matridigit[i][j] - other._matridigit[i][j] res[i][j] %= Matrix.mod return Matrix(res) def __mul__(self, other): if type(other) != int: if self.column != other.row: raise SizeError('sizes of matrixes are different') res = [[0 for j in range(other.column)] for i in range(self.row)] for i in range(self.row): for j in range(other.column): temp = 0 for k in range(self.column): temp += self._matridigit[i][k] * other._matrix[k][j] res[i][j] = temp % Matrix.mod return Matrix(res) else: n = other res = [[n * self._matridigit[i][j] % Matrix.mod for j in range(self.column)] for i in range(self.row)] return Matrix(res) def __pow__(self, m): if self.column != self.row: raise MatrixPowError('the size of row must be the same as that of column') n = self.row res = Matrix([[int(i == j) for i in range(n)] for j in range(n)]) while m: if m % 2 == 1: res = res * self self = self * self m //= 2 return res def __str__(self): res = [] for i in range(self.row): for j in range(self.column): res.append(str(self._matridigit[i][j])) res.append(' ') res.append('\n') res = res[:len(res) - 1] return ''.join(res) from collections import deque class Dinic: def __init__(self, N): self.N = N self.G = [[] for i in range(N)] def add_edge(self, fr, to, cap): forward = [to, cap, None] forward[2] = backward = [fr, 0, forward] self.G[fr].append(forward) self.G[to].append(backward) def add_multi_edge(self, v1, v2, cap1, cap2): edge1 = [v2, cap1, None] edge1[2] = edge2 = [v1, cap2, edge1] self.G[v1].append(edge1) self.G[v2].append(edge2) def bfs(self, s, t): self.level = level = [None] * self.N deq = deque([s]) level[s] = 0 G = self.G while deq: v = deq.popleft() lv = level[v] + 1 for (w, cap, _) in G[v]: if cap and level[w] is None: level[w] = lv deq.append(w) return level[t] is not None def dfs(self, v, t, f): if v == t: return f level = self.level for e in self.it[v]: (w, cap, rev) = e if cap and level[v] < level[w]: d = self.dfs(w, t, min(f, cap)) if d: e[1] -= d rev[1] += d return d return 0 def flow(self, s, t): flow = 0 INF = 10 ** 9 + 7 G = self.G while self.bfs(s, t): (*self.it,) = map(iter, self.G) f = INF while f: f = self.dfs(s, t, INF) flow += f return flow import sys, random, bisect from collections import deque, defaultdict from heapq import heapify, heappop, heappush from itertools import permutations from math import gcd, log input = lambda : sys.stdin.readline().rstrip() mi = lambda : map(int, input().split()) li = lambda : list(mi()) N = int(input()) A = li() dp = [[False for r in range(N)] for l in range(N)] for l in range(N): tmp = [A[l]] dp[l][l] = True for r in range(l + 1, N): val = A[r] while tmp and tmp[-1] == val: val = tmp[-1] + 1 tmp.pop() tmp.append(val) if len(tmp) == 1: dp[l][r] = True res = [i for i in range(N + 1)] for r in range(1, N + 1): for l in range(1, r + 1): if dp[l - 1][r - 1]: res[r] = min(res[r], 1 + res[l - 1]) print(res[N])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

from sys import stdin, gettrace if not gettrace(): def input(): return next(stdin)[:-1] INF = 10000 def main(): n = int(input()) aa = [int(a) for a in input().split()] dp = [[0] * (n + 1) for _ in range(n)] def calc_dp(i, j): if i + 1 == j: dp[i][j] = aa[i] if dp[i][j] != 0: return dp[i][j] dp[i][j] = -1 for k in range(i + 1, j): lf = calc_dp(i, k) rg = calc_dp(k, j) if lf > 0 and lf == rg: dp[i][j] = lf + 1 break return dp[i][j] dp2 = list(range(0, n + 1)) for i in range(n): for j in range(i + 1, n + 1): if calc_dp(i, j) > 0: dp2[j] = min(dp2[j], dp2[i] + 1) print(dp2[n]) main()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) a = list(map(int, input().split(' '))) new_a = [[0] * 600 for i in range(600)] dp = [[2147483647] * 600 for i in range(600)] for i in range(n): new_a[i + 1][i + 1] = a[i] dp[i + 1][i + 1] = 1 for i in range(1, n + 1): for j in range(i + 1, n + 1): dp[i][j] = j - i + 1 for llen in range(2, n + 1): for left in range(1, n - llen + 2): right = left + llen - 1 for middle in range(left, right): dp[left][right] = min(dp[left][right], dp[left][middle] + dp[middle + 1][right]) if dp[left][middle] == 1 and dp[middle + 1][right] == 1 and (new_a[left][middle] == new_a[middle + 1][right]): dp[left][right] = 1 new_a[left][right] = new_a[left][middle] + 1 print(dp[1][n])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import os import sys from io import BytesIO, IOBase import heapq as h from bisect import bisect_left, bisect_right from types import GeneratorType BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): import os self.os = os self._fd = file.fileno() self.buffer = BytesIO() self.writable = 'x' in file.mode or 'r' not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b'\n') + (not b) ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: self.os.write(self._fd, self.buffer.getvalue()) (self.buffer.truncate(0), self.buffer.seek(0)) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode('ascii')) self.read = lambda : self.buffer.read().decode('ascii') self.readline = lambda : self.buffer.readline().decode('ascii') (sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout)) input = lambda : sys.stdin.readline().rstrip('\r\n') from collections import defaultdict as dd, deque as dq import math, string def getInts(): return [int(s) for s in input().split()] def getInt(): return int(input()) def getStrs(): return [s for s in input().split()] def getStr(): return input() def listStr(): return list(input()) MOD = 998244353 def solve(): N = getInt() A = getInts() dp = [[-1 for j in range(N)] for i in range(N)] for i in range(N): dp[i][i] = A[i] for X in range(2, N + 1): for i in range(N - X + 1): j = i + X - 1 for k in range(i, j): if dp[i][k] == dp[k + 1][j] and dp[i][k] != -1: dp[i][j] = dp[i][k] + 1 break ans = [10 ** 9 + 1] * (N + 1) ans[0] = 0 for i in range(1, N + 1): for k in range(1, i + 1): if dp[k - 1][i - 1] != -1: ans[i] = min(ans[i], ans[k - 1] + 1) return ans[N] print(solve())

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import os from io import BytesIO, IOBase import sys from collections import defaultdict, deque, Counter from math import sqrt, pi, ceil, log, inf, gcd, floor from itertools import combinations, permutations from bisect import * from fractions import Fraction from heapq import * from random import randint def main(): n = int(input()) a = list(map(int, input().split())) dp = [[0 for _ in range(n)] for _ in range(n)] for i in range(n): dp[i][i] = a[i] for i in range(n - 2, -1, -1): for j in range(i + 1, n, 1): for k in range(i, j, 1): if dp[i][k] and dp[i][k] == dp[k + 1][j]: dp[i][j] = dp[i][k] + 1 b = [10 ** 10] * (n + 1) b[0] = 0 for i in range(1, n + 1): for j in range(i): if dp[j][i - 1]: b[i] = min(b[i], b[j] + 1) print(b[n]) BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = 'x' in file.mode or 'r' not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b'\n') + (not b) ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) (self.buffer.truncate(0), self.buffer.seek(0)) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode('ascii')) self.read = lambda : self.buffer.read().decode('ascii') self.readline = lambda : self.buffer.readline().decode('ascii') (sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout)) input = lambda : sys.stdin.readline().rstrip('\r\n') main()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

def f(): n = int(input()) A = [int(s) for s in input().split()] memo = [[None for j in range(n + 1)] for i in range(n + 1)] for i in range(n): memo[i][i] = [A[i], A[i], 1] for l in range(2, n + 1): for left in range(0, n - l + 1): right = left + l - 1 minLen = l shortestMid = right for mid in range(left + 1, right + 1): pre = memo[left][mid - 1] post = memo[mid][right] combLen = pre[2] + post[2] if pre[1] == post[0]: combLen -= 1 if combLen < minLen: minLen = combLen shortestMid = mid pre = memo[left][shortestMid - 1] post = memo[shortestMid][right] startEle = pre[0] endEle = post[1] if pre[2] == 1: if pre[0] == post[0]: startEle = pre[0] + 1 if post[2] == 1: if pre[1] == post[0]: endEle = post[0] + 1 memo[left][right] = [startEle, endEle, minLen] print(memo[0][n - 1][2]) f()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys import bisect import heapq from collections import defaultdict as dd from collections import deque from collections import Counter as c from itertools import combinations as comb from bisect import bisect_left as bl, bisect_right as br, bisect mod = pow(10, 9) + 7 mod2 = 998244353 def data(): return sys.stdin.readline().strip() def out(var): sys.stdout.write(var) def l(): return list(map(int, data().split())) def sl(): return list(map(str, data().split())) def sp(): return map(int, data().split()) def ssp(): return map(str, data().split()) def l1d(n, val=0): return [val for i in range(n)] def l2d(n, m, val=0): return [[val for i in range(n)] for j in range(m)] n = int(data()) arr = l() dp = [[0 for j in range(500)] for i in range(500)] dp2 = [0 for i in range(501)] for i in range(n): dp[i][i] = arr[i] i = n - 2 while ~i: j = i + 1 while j < n: dp[i][j] = -1 for k in range(i, j): if (~dp[i][k] and dp[i][k]) == dp[k + 1][j]: dp[i][j] = dp[i][k] + 1 j += 1 i -= 1 for i in range(1, n + 1): dp2[i] = pow(10, 9) for j in range(i): if ~dp[j][i - 1]: dp2[i] = min(dp2[i], dp2[j] + 1) out(str(dp2[n]))

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

from heapq import * import sys int1 = lambda x: int(x) - 1 p2D = lambda x: print(*x, sep='\n') def II(): return int(sys.stdin.readline()) def MI(): return map(int, sys.stdin.readline().split()) def LI(): return list(map(int, sys.stdin.readline().split())) def LLI(rows_number): return [LI() for _ in range(rows_number)] def SI(): return sys.stdin.readline()[:-1] def main(): inf = 10 ** 9 n = II() aa = LI() dp1 = [[-1] * (n + 1) for _ in range(n)] to = [[i + 1] for i in range(n)] for i in range(n): dp1[i][i + 1] = aa[i] for w in range(2, n + 1): for l in range(n - w + 1): r = l + w for m in range(l + 1, r): if dp1[l][m] != -1 and dp1[l][m] == dp1[m][r]: dp1[l][r] = dp1[l][m] + 1 to[l].append(r) hp = [] heappush(hp, (0, 0)) dist = [-1] * (n + 1) while hp: (d, i) = heappop(hp) if i == n: print(d) break if dist[i] != -1: continue dist[i] = d for j in to[i]: if dist[j] != -1: continue heappush(hp, (d + 1, j)) main()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys from array import array from typing import List, Tuple, TypeVar, Generic, Sequence, Union def input(): return sys.stdin.buffer.readline().decode('utf-8') def main(): n = int(input()) a = list(map(int, input().split())) dp = [array('h', [10000]) * (n + 1) for _ in range(n + 1)] num = [array('h', [-1]) * (n + 1) for _ in range(n + 1)] for i in range(n): dp[i][i + 1] = 1 num[i][i + 1] = a[i] for sublen in range(2, n + 1): for (l, r) in zip(range(n), range(sublen, n + 1)): for mid in range(l + 1, r): if num[l][mid] == num[mid][r] != -1: dp[l][r] = 1 num[l][r] = num[l][mid] + 1 break dp[l][r] = min(dp[l][r], dp[l][mid] + dp[mid][r]) print(dp[0][-1]) main()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) a = [int(x) for x in input().split()] dp = [[-1] * (n + 1) for _ in range(n + 1)] for i in range(n): dp[i][i + 1] = a[i] for leng in range(2, n + 1): for l in range(n + 1): if l + leng > n: continue r = l + leng for mid in range(l + 1, n + 1): if dp[l][mid] != -1 and dp[l][mid] == dp[mid][r]: dp[l][r] = dp[l][mid] + 1 dp2 = [float('inf') for _ in range(n + 1)] for i in range(n + 1): dp2[i] = i for j in range(i): if dp[j][i] != -1: dp2[i] = min(dp2[i], dp2[j] + 1) print(dp2[n])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys input = sys.stdin.readline n = int(input().strip()) a = [int(x) for x in input().strip().split()] dp = [[0] * n for i in range(n)] for i in range(n): dp[i][i] = [a[i], 1] for i in range(1, n): for j in range(n - i): (v, c) = (-1, i + 1) for k in range(i): if dp[j][j + k][0] != -1 and dp[j][j + k][0] == dp[j + k + 1][j + i][0]: (v, c) = (dp[j][j + k][0] + 1, 1) break else: (v, c) = (-1, min(c, dp[j][j + k][1] + dp[j + k + 1][j + i][1])) dp[j][j + i] = [v, c] print(dp[0][-1][1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys pl = 1 if pl: input = sys.stdin.readline else: sys.stdin = open('input.txt', 'r') sys.stdout = open('outpt.txt', 'w') def li(): return [int(xxx) for xxx in input().split()] def fi(): return int(input()) def si(): return list(input().rstrip()) def mi(): return map(int, input().split()) t = 1 while t > 0: t -= 1 n = fi() a = li() dp = [[0] * (n + 1) for i in range(n + 1)] for i in range(n - 1, -1, -1): for j in range(i, n): if i == j: dp[i][j] = a[i] elif i == j - 1: if a[i] == a[j]: dp[i][j] = a[i] + 1 else: for k in range(i, j): if dp[i][k] and dp[k + 1][j] and (dp[i][k] == dp[k + 1][j]): dp[i][j] = dp[i][k] + 1 break ans = [10 ** 18] * (n + 1) ans[-1] = 0 for i in range(n - 1, -1, -1): for j in range(i, n): if dp[i][j]: ans[i] = min(ans[i], 1 + ans[j + 1]) else: ans[i] = min(ans[i], j - i + 1 + ans[j + 1]) print(ans[0])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

from sys import stdin, stdout import sys n = int(stdin.readline().strip()) arr = list(map(int, stdin.readline().strip().split(' '))) dp_arr = [[None for i in range(n)] for i in range(n)] for i in range(n): dp_arr[i][i] = (arr[i], 1, arr[i]) def merge_small(c1, c2): if c1[1] == 1 and c2[1] == 1: if c1[0] == c2[0]: return (c1[0] + 1, 1, c1[0] + 1) else: return (c1[0], 2, c2[0]) elif c1[1] == 2 and c2[1] == 1: if c1[2] == c2[0]: if c1[0] == c1[2] + 1: return (c1[0] + 1, 1, c1[0] + 1) else: return (c1[0], 2, c2[2] + 1) else: return (c1[0], 3, c2[2]) elif c1[1] == 1 and c2[1] == 2: if c1[2] == c2[0]: if c2[2] == c2[0] + 1: return (c2[2] + 1, 1, c2[2] + 1) else: return (c2[0] + 1, 2, c2[2]) else: return (c1[0], 3, c2[2]) elif c1[1] == 2 and c2[1] == 2: if c1[2] == c2[0]: c1 = (c1[0], 2, c1[2] + 1) c2 = (c2[2], 1, c2[2]) if c1[1] == 2 and c2[1] == 1: if c1[2] == c2[0]: if c1[0] == c1[2] + 1: return (c1[0] + 1, 1, c1[0] + 1) else: return (c1[0], 2, c2[2] + 1) else: return (c1[0], 3, c2[2]) else: return (c1[0], 4, c2[2]) def merge_main(c1, c2): if c1[1] > 2: if c2[1] > 2: if c1[2] == c2[0]: return (c1[0], c1[1] + c2[1] - 1, c2[2]) else: return (c1[0], c1[1] + c2[1], c2[2]) else: if c2[1] == 1: if c1[2] == c2[0]: return (c1[0], c1[1], c2[2] + 1) else: return (c1[0], c1[1] + 1, c2[2]) if c2[1] == 2: if c1[2] == c2[0]: if c1[2] + 1 == c2[2]: return (c1[0], c1[1], c2[2] + 1) else: return (c1[0], c1[1] + 1, c2[2]) else: return (c1[0], c1[1] + 2, c2[2]) elif c2[1] > 2: if c1[1] == 1: if c1[2] == c2[0]: return (c1[2] + 1, c2[1], c2[2]) else: return (c1[2], c2[1] + 1, c2[2]) if c1[1] == 2: if c1[2] == c2[0]: if c1[0] == c1[2] + 1: return (c1[0] + 1, c2[1], c2[2]) else: return (c1[0], c2[1] + 1, c2[2]) else: return (c1[0], c2[1] + 2, c2[2]) else: return merge_small(c1, c2) for i1 in range(1, n): for j1 in range(n - i1): curr_pos = (j1, j1 + i1) for k1 in range(j1, j1 + i1): res = merge_main(dp_arr[j1][k1], dp_arr[k1 + 1][j1 + i1]) if dp_arr[j1][j1 + i1] == None or dp_arr[j1][j1 + i1][1] > res[1]: dp_arr[j1][j1 + i1] = res stdout.write(str(dp_arr[0][n - 1][1]) + '\n')

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys int1 = lambda x: int(x) - 1 p2D = lambda x: print(*x, sep='\n') def II(): return int(sys.stdin.readline()) def MI(): return map(int, sys.stdin.readline().split()) def LI(): return list(map(int, sys.stdin.readline().split())) def LLI(rows_number): return [LI() for _ in range(rows_number)] def SI(): return sys.stdin.readline()[:-1] def main(): inf = 10 ** 9 n = II() aa = LI() dp1 = [[-1] * n for _ in range(n)] dp2 = [[inf] * n for _ in range(n)] for i in range(n): dp1[i][i] = aa[i] dp2[i][i] = 1 for w in range(2, n + 1): for l in range(n - w + 1): r = l + w - 1 for m in range(l, r): if dp1[l][m] != -1 and dp1[l][m] == dp1[m + 1][r]: dp1[l][r] = dp1[l][m] + 1 dp2[l][r] = 1 for m in range(n): for l in range(m + 1): for r in range(m + 1, n): dp2[l][r] = min(dp2[l][r], dp2[l][m] + dp2[m + 1][r]) print(dp2[0][n - 1]) main()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import os import sys from io import BytesIO, IOBase BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = 'x' in file.mode or 'r' not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b'\n') + (not b) ptr = self.buffer.tell() (self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) (self.buffer.truncate(0), self.buffer.seek(0)) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode('ascii')) self.read = lambda : self.buffer.read().decode('ascii') self.readline = lambda : self.buffer.readline().decode('ascii') (sys.stdin, sys.stdout) = (IOWrapper(sys.stdin), IOWrapper(sys.stdout)) input = lambda : sys.stdin.readline().rstrip('\r\n') from math import factorial from collections import Counter, defaultdict, deque from heapq import heapify, heappop, heappush def RL(): return map(int, sys.stdin.readline().rstrip().split()) def RLL(): return list(map(int, sys.stdin.readline().rstrip().split())) def N(): return int(input()) def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0 def perm(n, m): return factorial(n) // factorial(n - m) if n >= m else 0 def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2) mod = 998244353 INF = float('inf') def main(): n = N() arr = RLL() dp = [[n] * n for i in range(n)] rec = [[0] * n for i in range(n)] for i in range(n): rec[i][i] = arr[i] dp[i][i] = 1 for le in range(2, n + 1): for l in range(n): r = l + le - 1 if r > n - 1: break for m in range(l, r): dp[l][r] = min(dp[l][r], dp[l][m] + dp[m + 1][r]) if rec[l][m] == rec[m + 1][r] and dp[l][m] == dp[m + 1][r] == 1: dp[l][r] = 1 rec[l][r] = rec[l][m] + 1 print(dp[0][-1]) main()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

from sys import stdin, stdout def dfs(l, r, dp, a_a): if l == r: return a_a[l] if l + 1 == r: if a_a[l] == a_a[r]: return a_a[l] + 1 else: return -1 if dp[l][r] != 10 ** 6: return dp[l][r] dp[l][r] = -1 for m in range(l, r): r1 = dfs(l, m, dp, a_a) r2 = dfs(m + 1, r, dp, a_a) if r1 > 0 and r1 == r2: dp[l][r] = r1 + 1 return dp[l][r] return dp[l][r] def array_shrinking(n, a_a): dp = [[10 ** 6 for _ in range(n)] for _ in range(n)] dp2 = [10 ** 6 for _ in range(n)] for i in range(n): dp2[i] = min(i + 1, dp2[i]) for j in range(i, n): r = dfs(i, j, dp, a_a) if r != -1: if i > 0: dp2[j] = min(dp2[i - 1] + 1, dp2[j]) else: dp2[j] = min(1, dp2[j]) return dp2[n - 1] n = int(stdin.readline()) a_a = list(map(int, stdin.readline().split())) res = array_shrinking(n, a_a) stdout.write(str(res))

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys, math import io, os from bisect import bisect_left as bl, bisect_right as br, insort from collections import defaultdict as dd, deque, Counter def data(): return sys.stdin.readline().strip() def mdata(): return list(map(int, data().split())) def outl(var): sys.stdout.write(' '.join(map(str, var)) + '\n') def out(var): sys.stdout.write(str(var) + '\n') from decimal import Decimal INF = 10001 mod = int(1000000000.0) + 7 n = int(data()) a = mdata() ans = [n] dp1 = [[0] * n for i in range(n)] dp2 = [[n] * n for i in range(n)] for i in range(n - 1, -1, -1): dp1[i][i] = a[i] dp2[i][i] = 1 for j in range(i + 1, n): for k in range(i, j): if dp1[i][k] == dp1[k + 1][j] != 0: dp1[i][j] = dp1[i][k] + 1 dp2[i][j] = 1 dp2[i][j] = min(dp2[i][j], dp2[i][k] + dp2[k + 1][j]) out(dp2[0][n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import io import os import sys from functools import lru_cache from collections import defaultdict sys.setrecursionlimit(10 ** 5) def solve(N, A): valToLeftRight = defaultdict(lambda : defaultdict(set)) valToRightLeft = defaultdict(lambda : defaultdict(set)) for (i, x) in enumerate(A): valToLeftRight[x][i].add(i) valToRightLeft[x][i].add(i) maxVal = 1000 + 10 for val in range(maxVal): for (l, rights) in valToLeftRight[val - 1].items(): for r in rights: l2 = r + 1 if l2 in valToLeftRight[val - 1]: for r2 in valToLeftRight[val - 1][l2]: assert l <= r assert r + 1 == l2 assert l2 <= r2 valToLeftRight[val][l].add(r2) valToRightLeft[val][r2].add(l) r2 = l - 1 if r2 in valToRightLeft[val - 1]: for l2 in valToRightLeft[val - 1][r2]: assert l2 <= r2 assert r2 == l - 1 assert l <= r valToLeftRight[val][l2].add(r) valToRightLeft[val][r].add(l2) intervals = defaultdict(list) for val in range(maxVal): for (l, rights) in valToLeftRight[val].items(): for r in rights: intervals[l].append(r) @lru_cache(maxsize=None) def getBest(left): if left == N: return 0 best = float('inf') for right in intervals[left]: best = min(best, 1 + getBest(right + 1)) return best return getBest(0) def tup(l, r): return l * 16384 + r def untup(t): return divmod(t, 16384) def solve(N, A): cache = {} def f(lr): if lr not in cache: (l, r) = untup(lr) if r - l == 1: return tup(1, A[l]) best = tup(float('inf'), float('inf')) for i in range(l + 1, r): lSplit = f(tup(l, i)) rSplit = f(tup(i, r)) (lLen, lVal) = untup(lSplit) (rLen, rVal) = untup(rSplit) if lLen != 1 or rLen != 1: best = min(best, tup(lLen + rLen, 9999)) elif lVal == rVal: best = min(best, tup(1, lVal + 1)) else: best = min(best, tup(2, 9999)) cache[lr] = best return cache[lr] ans = untup(f(tup(0, N)))[0] return ans input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline (N,) = list(map(int, input().split())) A = list(map(int, input().split())) ans = solve(N, A) print(ans)

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

def rr(): return input().rstrip() def rri(): return int(rr()) def rrm(): return list(map(int, rr().split())) from collections import defaultdict def mus(d=0): return defaultdict(defaultdict(d)) def ms(x, y, d=0): return [[d] * y for i in range(x)] def ar(x, d=0): return [d] * x def ppm(m, n=0, x=0, y=0): print('\n'.join(('\t'.join((str(m[j][i]) for j in range(y or n))) for i in range(x or n)))) def ppa(a, n): print('\t'.join(map(str, a[0:n]))) def ppl(): print('\n+' + '- -' * 20 + '+\n') INF = float('inf') def fake_input(): return ... dp = ms(501, 501) dp2 = ar(501, INF) def read(): n = rri() global arr arr = rrm() return (arr, n) def calc_dp(l, r): assert l < r if l + 1 == r: dp[l][r] = arr[l] return dp[l][r] if dp[l][r] != 0: return dp[l][r] dp[l][r] = -1 for i in range(l + 1, r): lf = calc_dp(l, i) rg = calc_dp(i, r) if lf > 0 and lf == rg: dp[l][r] = lf + 1 return dp[l][r] return dp[l][r] def solve(arr, n): dp2[0] = 0 for i in range(n): for j in range(i + 1, n + 1): v = calc_dp(i, j) if v > 0: dp2[j] = min(dp2[j], dp2[i] + 1) ans = dp2[n] return ans input_data = read() result = solve(*input_data) print(result)

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) li = list(map(int, input().split(' '))) dp1 = [] for i in range(n): lis = [-1] * n dp1.append(lis) dp2 = [0] * n for i in range(n): dp1[i][i] = li[i] for i in range(n): dp2[i] = i + 1 size = 2 while size <= n: i = 0 while i < n - size + 1: j = i + size - 1 k = i while k < j: if dp1[i][k] != -1: if dp1[i][k] == dp1[k + 1][j]: dp1[i][j] = dp1[i][k] + 1 k += 1 i += 1 size += 1 i = 0 while i < n: k = 0 while k <= i: if dp1[k][i] != -1: if k == 0: dp2[i] = 1 else: dp2[i] = min(dp2[i], dp2[k - 1] + 1) k += 1 i += 1 print(dp2[n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import math input_list = lambda : list(map(int, input().split())) n = int(input()) a = input_list() (rows, cols) = (n + 1, n + 1) dp = [[-1 for i in range(rows)] for j in range(cols)] for i in range(n): dp[i][i] = a[i] for last in range(1, n): for first in range(last - 1, -1, -1): for mid in range(last, first, -1): if dp[first][mid - 1] != -1 and dp[mid][last] != -1 and (dp[first][mid - 1] == dp[mid][last]): dp[first][last] = dp[first][mid - 1] + 1 ans = [0 for i in range(n)] for i in range(n): ans[i] = i + 1 for i in range(n): for j in range(i, -1, -1): if j - 1 >= 0: if dp[j][i] != -1: ans[i] = min(ans[i], ans[j - 1] + 1) elif dp[0][i] != -1: ans[i] = 1 print(ans[n - 1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) b = [int(_) for _ in input().split()] d = [[b[i] if i == j else -1 for i in range(n)] for j in range(n)] def f(i, j): if d[i][j] != -1: return d[i][j] d[i][j] = 0 for m in range(i, j): l = f(i, m) if f(m + 1, j) == l and l: d[i][j] = l + 1 break return d[i][j] a = [_ for _ in range(1, n + 1)] for e in range(1, n): for s in range(e + 1): if f(s, e): a[e] = min(a[e], a[s - 1] + 1 if s > 0 else a[s]) print(a[-1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

def examA(): T = I() ans = [] for _ in range(T): (N, M) = LI() if N % M != 0: ans.append('NO') else: ans.append('YES') for v in ans: print(v) return def examB(): T = I() ans = [] for _ in range(T): N = I() A = LI() A.sort() ans.append(A[::-1]) for v in ans: print(' '.join(map(str, v))) return def examC(): T = I() ans = [] for _ in range(T): (N, K) = LI() A = LI() sumA = sum(A) if sumA == 0: ans.append('YES') continue cur = 0 L = [] for i in range(100): now = K ** i L.append(now) cur += now if cur >= sumA: break for i in range(N): A[i] *= -1 heapify(A) for l in L[::-1]: if not A: break a = -heappop(A) if a < l: heappush(A, -a) elif a > l: heappush(A, -(a - l)) if not A or heappop(A) == 0: ans.append('YES') else: ans.append('NO') for v in ans: print(v) return def examD(): class combination: def __init__(self, n, mod): self.n = n self.fac = [1] * (n + 1) self.inv = [1] * (n + 1) for j in range(1, n + 1): self.fac[j] = self.fac[j - 1] * j % mod self.inv[n] = pow(self.fac[n], mod - 2, mod) for j in range(n - 1, -1, -1): self.inv[j] = self.inv[j + 1] * (j + 1) % mod def comb(self, n, r, mod): if r > n or n < 0 or r < 0: return 0 return self.fac[n] * self.inv[n - r] * self.inv[r] % mod (N, M) = LI() ans = 0 if N == 2: print(ans) return C = combination(M, mod2) for i in range(N - 1, M + 1): cur = pow(2, N - 3, mod2) * (i - 1) * C.comb(i - 2, N - 3, mod2) ans += cur ans %= mod2 print(ans) return def examE(): N = I() A = LI() dp = [[-1] * (N + 1) for _ in range(N + 1)] for i in range(N): dp[i][i + 1] = A[i] for l in range(2, N + 1): for i in range(N - l + 1): for k in range(i + 1, i + l): if dp[i][k] >= 1 and dp[i][k] == dp[k][i + l]: dp[i][i + l] = dp[i][k] + 1 L = [inf] * (N + 1) for i in range(1, N + 1): if dp[0][i] >= 1: L[i] = 1 for i in range(N): for k in range(1, N - i + 1): if dp[i][i + k] >= 1: L[i + k] = min(L[i + k], L[i] + 1) ans = L[N] print(ans) return def examF(): ans = 0 print(ans) return import sys, copy, bisect, itertools, heapq, math, random from heapq import heappop, heappush, heapify from collections import Counter, defaultdict, deque def I(): return int(sys.stdin.readline()) def LI(): return list(map(int, sys.stdin.readline().split())) def LSI(): return list(map(str, sys.stdin.readline().split())) def LS(): return sys.stdin.readline().split() def SI(): return sys.stdin.readline().strip() global mod, mod2, inf, alphabet, _ep mod = 10 ** 9 + 7 mod2 = 998244353 inf = 10 ** 18 _ep = 10 ** (-12) alphabet = [chr(ord('a') + i) for i in range(26)] examE()

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

n = int(input()) b = [int(_) for _ in input().split()] d = [[-1 if i != j else b[i] for i in range(n)] for j in range(n)] for l in range(1, n): for s in range(n - l): e = s + l for m in range(s, e): if d[s][m] == d[m + 1][e] and d[s][m] != -1: d[s][e] = d[s][m] + 1 a = [1] for e in range(1, n): t = 4096 for s in range(e + 1): if d[s][e] != -1: t = min(t, a[s - 1] + 1 if s > 0 else a[s]) a.append(t) print(a[-1])

python

You are given an array $a_1, a_2, \dots, a_n$. You can perform the following operation any number of times: Choose a pair of two neighboring equal elements $a_i = a_{i + 1}$ (if there is at least one such pair). Replace them by one element with value $a_i + 1$. After each such operation, the length of the array will decrease by one (and elements are renumerated accordingly). What is the minimum possible length of the array $a$ you can get? -----Input----- The first line contains the single integer $n$ ($1 \le n \le 500$) — the initial length of the array $a$. The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 1000$) — the initial array $a$. -----Output----- Print the only integer — the minimum possible length you can get after performing the operation described above any number of times. -----Examples----- Input 5 4 3 2 2 3 Output 2 Input 7 3 3 4 4 4 3 3 Output 2 Input 3 1 3 5 Output 3 Input 1 1000 Output 1 -----Note----- In the first test, this is one of the optimal sequences of operations: $4$ $3$ $2$ $2$ $3$ $\rightarrow$ $4$ $3$ $3$ $3$ $\rightarrow$ $4$ $4$ $3$ $\rightarrow$ $5$ $3$. In the second test, this is one of the optimal sequences of operations: $3$ $3$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $3$ $3$ $\rightarrow$ $4$ $4$ $4$ $4$ $4$ $\rightarrow$ $5$ $4$ $4$ $4$ $\rightarrow$ $5$ $5$ $4$ $\rightarrow$ $6$ $4$. In the third and fourth tests, you can't perform the operation at all.

taco

import sys def GRIG(L): LENT = len(L) MINT = 1 GOT = 0 DY = [[{x: 0 for x in range(0, 10)}, 0, 0]] for i in L: DY.append([{x: 0 for x in range(0, 10)}, 0, 0]) GOT += 1 for j in range(0, GOT): if DY[j][0][i] == 1: DY[j][0][i] = 0 DY[j][1] -= 1 else: DY[j][0][i] = 1 DY[j][1] += 1 DY[j][2] += 1 if DY[j][1] <= 1 and DY[j][2] > MINT: MINT = DY[j][2] return MINT TESTCASES = int(input().strip()) for i in range(0, TESTCASES): L = [int(x) for x in list(input().strip())] print(GRIG(L))

python

There are players standing in a row each player has a digit written on their T-Shirt (multiple players can have the same number written on their T-Shirt). You have to select a group of players, note that players in this group should be standing in $\textbf{consecutive fashion}$. For example second player of chosen group next to first player of chosen group, third player next to second and similarly last player next to second last player of chosen group. Basically You've to choose a contiguous group of players. After choosing a group, players can be paired if they have the same T-Shirt number (one player can be present in at most one pair), finally the chosen group is called “good” if at most one player is left unmatched. Your task is to find the size of the maximum “good” group. Formally, you are given a string $S=s_{1}s_{2}s_{3}...s_{i}...s_{n}$ where $s_{i}$ can be any digit character between $'0'$ and $'9'$ and $s_{i}$ denotes the number written on the T-Shirt of $i^{th}$ player. Find a value $length$ such that there exist pair of indices $(i,j)$ which denotes $S[i...j]$ is a “good” group where $i\geq1$ and $j\leq S.length$ and $i\leq j$ and $(j-i+1)=length$ and there exist no other pair $(i’,j’)$ such that $(j’-i’+1)>length$ and $S[i'...j']$ is a "good" group. -----Input:----- - First line will contain $T$, number of testcases. Then the testcases follow. - $i^{th}$ testcase consist of a single line of input, a string $S$. -----Output:----- For each testcase, output in a single line maximum possible size of a "good" group. -----Constraints----- $\textbf{Subtask 1} (20 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{3}$ $\textbf{Subtask 2} (80 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{5}$ -----Sample Input:----- 1 123343 -----Sample Output:----- 3 -----EXPLANATION:----- 1$\textbf{$\underline{2 3 3}$}$43 Underlined group is a “good” group because the second player(number 2 on T-Shirt) is the only player who is left unmatched and third and fourth player can form a pair, no other group has length greater than 3 that are “good”. However note that we have other “good” group also 12$\textbf{$\underline{334}$}$3 but length is 3 which is same as our answer. -----Sample Input:----- 1 95665 -----Sample Output:----- 5 -----EXPLANATION:----- $\textbf{$\underline{95665}$}$ is “good” group because first player is the only player who is left unmatched second and fifth player can form pair and third and fourth player also form pair. -----Sample Input:----- 2 2323 1234567 -----Sample Output:----- 4 1 -----EXPLANATION:----- For first test case $\textbf{$\underline{2323}$}$ is a “good” group because there are no players who are left unmatched first and third player form pair and second and fourth player form pair. For second test Only length one "good" group is possible.

taco

import sys import math from collections import defaultdict, Counter def possible(n1, n): odd = set() for j in range(n): if s[j] in odd: odd.remove(s[j]) else: odd.add(s[j]) if j < n1 - 1: continue if len(odd) <= 1: return True if s[j - n1 + 1] in odd: odd.remove(s[j - n1 + 1]) else: odd.add(s[j - n1 + 1]) return False t = int(input()) for i in range(t): s = input().strip() n = len(s) ans = 0 for j in range(n, 0, -1): if possible(j, n): ans = j break print(ans)

python

There are players standing in a row each player has a digit written on their T-Shirt (multiple players can have the same number written on their T-Shirt). You have to select a group of players, note that players in this group should be standing in $\textbf{consecutive fashion}$. For example second player of chosen group next to first player of chosen group, third player next to second and similarly last player next to second last player of chosen group. Basically You've to choose a contiguous group of players. After choosing a group, players can be paired if they have the same T-Shirt number (one player can be present in at most one pair), finally the chosen group is called “good” if at most one player is left unmatched. Your task is to find the size of the maximum “good” group. Formally, you are given a string $S=s_{1}s_{2}s_{3}...s_{i}...s_{n}$ where $s_{i}$ can be any digit character between $'0'$ and $'9'$ and $s_{i}$ denotes the number written on the T-Shirt of $i^{th}$ player. Find a value $length$ such that there exist pair of indices $(i,j)$ which denotes $S[i...j]$ is a “good” group where $i\geq1$ and $j\leq S.length$ and $i\leq j$ and $(j-i+1)=length$ and there exist no other pair $(i’,j’)$ such that $(j’-i’+1)>length$ and $S[i'...j']$ is a "good" group. -----Input:----- - First line will contain $T$, number of testcases. Then the testcases follow. - $i^{th}$ testcase consist of a single line of input, a string $S$. -----Output:----- For each testcase, output in a single line maximum possible size of a "good" group. -----Constraints----- $\textbf{Subtask 1} (20 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{3}$ $\textbf{Subtask 2} (80 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{5}$ -----Sample Input:----- 1 123343 -----Sample Output:----- 3 -----EXPLANATION:----- 1$\textbf{$\underline{2 3 3}$}$43 Underlined group is a “good” group because the second player(number 2 on T-Shirt) is the only player who is left unmatched and third and fourth player can form a pair, no other group has length greater than 3 that are “good”. However note that we have other “good” group also 12$\textbf{$\underline{334}$}$3 but length is 3 which is same as our answer. -----Sample Input:----- 1 95665 -----Sample Output:----- 5 -----EXPLANATION:----- $\textbf{$\underline{95665}$}$ is “good” group because first player is the only player who is left unmatched second and fifth player can form pair and third and fourth player also form pair. -----Sample Input:----- 2 2323 1234567 -----Sample Output:----- 4 1 -----EXPLANATION:----- For first test case $\textbf{$\underline{2323}$}$ is a “good” group because there are no players who are left unmatched first and third player form pair and second and fourth player form pair. For second test Only length one "good" group is possible.

taco

from collections import defaultdict t = int(input()) for _ in range(t): s = input() s = list(s) l = s[:] n = len(s) ans = 0 if n <= 1000: if n <= 100: l = [] for i in range(n): for j in range(i + 1, n): l.append(s[i:j]) ans = 0 for i in l: dic = defaultdict(lambda : 0) for j in i: dic[j] += 1 cnt = 0 for k in dic.keys(): if dic[k] % 2 == 1: cnt += 1 if cnt <= 1: ans = max(ans, len(i)) print(ans) continue for i in range(n): dic = defaultdict(lambda : 0) unmatched = 0 for j in range(i, n): dic[l[j]] += 1 if dic[l[j]] % 2 == 1: unmatched += 1 else: unmatched -= 1 if unmatched <= 1: ans = max(ans, j - i + 1) print(ans) else: for i in range(n): dic = defaultdict(lambda : 0) unmatched = 0 for j in range(i + 1, n): dic[l[j]] += 1 if dic[l[j]] % 2 == 1: unmatched += 1 else: unmatched -= 1 if unmatched <= 1: ans = max(ans, j - i + 1) print(ans)

python

There are players standing in a row each player has a digit written on their T-Shirt (multiple players can have the same number written on their T-Shirt). You have to select a group of players, note that players in this group should be standing in $\textbf{consecutive fashion}$. For example second player of chosen group next to first player of chosen group, third player next to second and similarly last player next to second last player of chosen group. Basically You've to choose a contiguous group of players. After choosing a group, players can be paired if they have the same T-Shirt number (one player can be present in at most one pair), finally the chosen group is called “good” if at most one player is left unmatched. Your task is to find the size of the maximum “good” group. Formally, you are given a string $S=s_{1}s_{2}s_{3}...s_{i}...s_{n}$ where $s_{i}$ can be any digit character between $'0'$ and $'9'$ and $s_{i}$ denotes the number written on the T-Shirt of $i^{th}$ player. Find a value $length$ such that there exist pair of indices $(i,j)$ which denotes $S[i...j]$ is a “good” group where $i\geq1$ and $j\leq S.length$ and $i\leq j$ and $(j-i+1)=length$ and there exist no other pair $(i’,j’)$ such that $(j’-i’+1)>length$ and $S[i'...j']$ is a "good" group. -----Input:----- - First line will contain $T$, number of testcases. Then the testcases follow. - $i^{th}$ testcase consist of a single line of input, a string $S$. -----Output:----- For each testcase, output in a single line maximum possible size of a "good" group. -----Constraints----- $\textbf{Subtask 1} (20 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{3}$ $\textbf{Subtask 2} (80 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{5}$ -----Sample Input:----- 1 123343 -----Sample Output:----- 3 -----EXPLANATION:----- 1$\textbf{$\underline{2 3 3}$}$43 Underlined group is a “good” group because the second player(number 2 on T-Shirt) is the only player who is left unmatched and third and fourth player can form a pair, no other group has length greater than 3 that are “good”. However note that we have other “good” group also 12$\textbf{$\underline{334}$}$3 but length is 3 which is same as our answer. -----Sample Input:----- 1 95665 -----Sample Output:----- 5 -----EXPLANATION:----- $\textbf{$\underline{95665}$}$ is “good” group because first player is the only player who is left unmatched second and fifth player can form pair and third and fourth player also form pair. -----Sample Input:----- 2 2323 1234567 -----Sample Output:----- 4 1 -----EXPLANATION:----- For first test case $\textbf{$\underline{2323}$}$ is a “good” group because there are no players who are left unmatched first and third player form pair and second and fourth player form pair. For second test Only length one "good" group is possible.

taco

import sys def GRIG(L): MINT = 1 GOT = 0 DY = [[set(), 0, 0]] L_DY = 0 for i in L: L_DY += 1 DY.append([set(), 0, 0]) for j in range(L_DY - 1, -1, -1): if i in DY[j][0]: DY[j][0].remove(i) DY[j][1] -= 1 else: DY[j][0].add(i) DY[j][1] += 1 DY[j][2] += 1 if DY[j][2] > MINT and DY[j][1] <= 1: MINT = DY[j][2] return MINT TESTCASES = int(input().strip()) for i in range(0, TESTCASES): L = [int(x) for x in list(input().strip())] print(GRIG(L))

python

There are players standing in a row each player has a digit written on their T-Shirt (multiple players can have the same number written on their T-Shirt). You have to select a group of players, note that players in this group should be standing in $\textbf{consecutive fashion}$. For example second player of chosen group next to first player of chosen group, third player next to second and similarly last player next to second last player of chosen group. Basically You've to choose a contiguous group of players. After choosing a group, players can be paired if they have the same T-Shirt number (one player can be present in at most one pair), finally the chosen group is called “good” if at most one player is left unmatched. Your task is to find the size of the maximum “good” group. Formally, you are given a string $S=s_{1}s_{2}s_{3}...s_{i}...s_{n}$ where $s_{i}$ can be any digit character between $'0'$ and $'9'$ and $s_{i}$ denotes the number written on the T-Shirt of $i^{th}$ player. Find a value $length$ such that there exist pair of indices $(i,j)$ which denotes $S[i...j]$ is a “good” group where $i\geq1$ and $j\leq S.length$ and $i\leq j$ and $(j-i+1)=length$ and there exist no other pair $(i’,j’)$ such that $(j’-i’+1)>length$ and $S[i'...j']$ is a "good" group. -----Input:----- - First line will contain $T$, number of testcases. Then the testcases follow. - $i^{th}$ testcase consist of a single line of input, a string $S$. -----Output:----- For each testcase, output in a single line maximum possible size of a "good" group. -----Constraints----- $\textbf{Subtask 1} (20 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{3}$ $\textbf{Subtask 2} (80 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{5}$ -----Sample Input:----- 1 123343 -----Sample Output:----- 3 -----EXPLANATION:----- 1$\textbf{$\underline{2 3 3}$}$43 Underlined group is a “good” group because the second player(number 2 on T-Shirt) is the only player who is left unmatched and third and fourth player can form a pair, no other group has length greater than 3 that are “good”. However note that we have other “good” group also 12$\textbf{$\underline{334}$}$3 but length is 3 which is same as our answer. -----Sample Input:----- 1 95665 -----Sample Output:----- 5 -----EXPLANATION:----- $\textbf{$\underline{95665}$}$ is “good” group because first player is the only player who is left unmatched second and fifth player can form pair and third and fourth player also form pair. -----Sample Input:----- 2 2323 1234567 -----Sample Output:----- 4 1 -----EXPLANATION:----- For first test case $\textbf{$\underline{2323}$}$ is a “good” group because there are no players who are left unmatched first and third player form pair and second and fourth player form pair. For second test Only length one "good" group is possible.

taco

from sys import stdin for _ in range(int(stdin.readline())): s = stdin.readline().strip() m = 1 for i in range(len(s)): k = set() k.add(s[i]) for j in range(i + 1, len(s)): if s[j] in k: k.remove(s[j]) else: k.add(s[j]) if len(k) < 2: m = max(m, j - i + 1) print(m)

python

There are players standing in a row each player has a digit written on their T-Shirt (multiple players can have the same number written on their T-Shirt). You have to select a group of players, note that players in this group should be standing in $\textbf{consecutive fashion}$. For example second player of chosen group next to first player of chosen group, third player next to second and similarly last player next to second last player of chosen group. Basically You've to choose a contiguous group of players. After choosing a group, players can be paired if they have the same T-Shirt number (one player can be present in at most one pair), finally the chosen group is called “good” if at most one player is left unmatched. Your task is to find the size of the maximum “good” group. Formally, you are given a string $S=s_{1}s_{2}s_{3}...s_{i}...s_{n}$ where $s_{i}$ can be any digit character between $'0'$ and $'9'$ and $s_{i}$ denotes the number written on the T-Shirt of $i^{th}$ player. Find a value $length$ such that there exist pair of indices $(i,j)$ which denotes $S[i...j]$ is a “good” group where $i\geq1$ and $j\leq S.length$ and $i\leq j$ and $(j-i+1)=length$ and there exist no other pair $(i’,j’)$ such that $(j’-i’+1)>length$ and $S[i'...j']$ is a "good" group. -----Input:----- - First line will contain $T$, number of testcases. Then the testcases follow. - $i^{th}$ testcase consist of a single line of input, a string $S$. -----Output:----- For each testcase, output in a single line maximum possible size of a "good" group. -----Constraints----- $\textbf{Subtask 1} (20 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{3}$ $\textbf{Subtask 2} (80 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{5}$ -----Sample Input:----- 1 123343 -----Sample Output:----- 3 -----EXPLANATION:----- 1$\textbf{$\underline{2 3 3}$}$43 Underlined group is a “good” group because the second player(number 2 on T-Shirt) is the only player who is left unmatched and third and fourth player can form a pair, no other group has length greater than 3 that are “good”. However note that we have other “good” group also 12$\textbf{$\underline{334}$}$3 but length is 3 which is same as our answer. -----Sample Input:----- 1 95665 -----Sample Output:----- 5 -----EXPLANATION:----- $\textbf{$\underline{95665}$}$ is “good” group because first player is the only player who is left unmatched second and fifth player can form pair and third and fourth player also form pair. -----Sample Input:----- 2 2323 1234567 -----Sample Output:----- 4 1 -----EXPLANATION:----- For first test case $\textbf{$\underline{2323}$}$ is a “good” group because there are no players who are left unmatched first and third player form pair and second and fourth player form pair. For second test Only length one "good" group is possible.

taco

def isgood(s): c = 0 for i in set(s): k = s.count(i) if k % 2 and c: return 0 elif k % 2: c = 1 return 1 tc = int(input()) for i in range(tc): flag = 0 s = input() n = len(s) g = n while g: for i in range(n - g + 1): if isgood(s[i:i + g]): print(g) flag = 1 break g -= 1 if flag: break

python

There are players standing in a row each player has a digit written on their T-Shirt (multiple players can have the same number written on their T-Shirt). You have to select a group of players, note that players in this group should be standing in $\textbf{consecutive fashion}$. For example second player of chosen group next to first player of chosen group, third player next to second and similarly last player next to second last player of chosen group. Basically You've to choose a contiguous group of players. After choosing a group, players can be paired if they have the same T-Shirt number (one player can be present in at most one pair), finally the chosen group is called “good” if at most one player is left unmatched. Your task is to find the size of the maximum “good” group. Formally, you are given a string $S=s_{1}s_{2}s_{3}...s_{i}...s_{n}$ where $s_{i}$ can be any digit character between $'0'$ and $'9'$ and $s_{i}$ denotes the number written on the T-Shirt of $i^{th}$ player. Find a value $length$ such that there exist pair of indices $(i,j)$ which denotes $S[i...j]$ is a “good” group where $i\geq1$ and $j\leq S.length$ and $i\leq j$ and $(j-i+1)=length$ and there exist no other pair $(i’,j’)$ such that $(j’-i’+1)>length$ and $S[i'...j']$ is a "good" group. -----Input:----- - First line will contain $T$, number of testcases. Then the testcases follow. - $i^{th}$ testcase consist of a single line of input, a string $S$. -----Output:----- For each testcase, output in a single line maximum possible size of a "good" group. -----Constraints----- $\textbf{Subtask 1} (20 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{3}$ $\textbf{Subtask 2} (80 points)$ - $1 \leq T \leq 10$ - $S.length \leq 10^{5}$ -----Sample Input:----- 1 123343 -----Sample Output:----- 3 -----EXPLANATION:----- 1$\textbf{$\underline{2 3 3}$}$43 Underlined group is a “good” group because the second player(number 2 on T-Shirt) is the only player who is left unmatched and third and fourth player can form a pair, no other group has length greater than 3 that are “good”. However note that we have other “good” group also 12$\textbf{$\underline{334}$}$3 but length is 3 which is same as our answer. -----Sample Input:----- 1 95665 -----Sample Output:----- 5 -----EXPLANATION:----- $\textbf{$\underline{95665}$}$ is “good” group because first player is the only player who is left unmatched second and fifth player can form pair and third and fourth player also form pair. -----Sample Input:----- 2 2323 1234567 -----Sample Output:----- 4 1 -----EXPLANATION:----- For first test case $\textbf{$\underline{2323}$}$ is a “good” group because there are no players who are left unmatched first and third player form pair and second and fourth player form pair. For second test Only length one "good" group is possible.

taco

class Solution: def binTreeSortedLevels(self, arr, n): li = [] i = 0 level = 0 while i < n: dumm = [] if level == 0: li.append([arr[i]]) i += 1 level += 1 else: size = 2 ** level if i + size < n: dumm.extend(arr[i:i + size]) dumm.sort() li.append(dumm) i += size level += 1 else: dumm.extend(arr[i:]) dumm.sort() li.append(dumm) break return li

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): ans = [] m = 1 level = [] j = 0 for i in range(n): if j < m: level.append(arr[i]) j += 1 else: level.sort() ans.append(level.copy()) level.clear() m += m j = 1 level.append(arr[i]) level.sort() ans.append(level) return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): res = [] i = 0 ls = 1 while i < n: t = (1 << ls) - 1 t = min(t, n) temp = sorted(arr[i:t]) i = t ls += 1 res.append(temp) return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): res = [] (i, total) = (0, 0) while total < n: temp = [] for j in range(2 ** i): if total < n: temp.append(arr[total]) total += 1 else: break temp.sort() res.append(temp) i += 1 return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): n = len(arr) list2 = [[arr[0]]] c = 0 j = 1 list3 = [] for x in range(1, n): if c == 2 ** j - 1: list3.append(arr[x]) list3.sort() list2.append(list3) list3 = [] j += 1 c = 0 else: list3.append(arr[x]) c += 1 if len(list3) != 0: list3.sort() list2.append(list3) return list2

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

from collections import deque class Solution: def binTreeSortedLevels(self, arr, n): dq = deque() dq.append(0) res = [] while len(dq) > 0: currsize = len(dq) t = [] for i in range(currsize): temp = dq.popleft() t.append(arr[temp]) if 2 * temp + 1 < n: dq.append(2 * temp + 1) if 2 * temp + 2 < n: dq.append(2 * temp + 2) t.sort() res.append(t) return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): final = [] l = [1] final.append([arr[0]]) i = 0 while True: li = len(l) l = [] for j in range(li): if 2 * i + 1 < n: l.append(arr[2 * i + 1]) if 2 * i + 2 < n: l.append(arr[2 * i + 2]) i += 1 if len(l): final.append(sorted(l)) else: break return final

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): output = [] i = 0 while 2 ** i <= n: j = 2 ** i k = 2 ** (i + 1) i += 1 output.append(sorted(arr[j - 1:k - 1])) return output

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): a1 = {} queue = [0] queue1 = [0] while len(queue) > 0: x = queue.pop(0) y = queue1.pop(0) if y not in a1: a1[y] = [] a1[y].append(arr[x]) if 2 * x + 1 < len(arr): queue.append(2 * x + 1) queue1.append(y + 1) if 2 * x + 2 < len(arr): queue.append(2 * x + 2) queue1.append(y + 1) e = [] for i in range(max(a1) + 1): e.append(sorted(a1[i])) return e

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

from collections import deque from sortedcontainers import SortedList class Solution: def binTreeSortedLevels(self, arr, n): q = deque([0]) res = [] while q: t = SortedList() for _ in range(len(q)): cur = q.popleft() t.add(arr[cur]) for i in [2 * cur + 1, 2 * cur + 2]: if i < len(arr): q.append(i) res.append(t) return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

from collections import deque from heapq import heappush, heappop class Solution: def binTreeSortedLevels(self, arr, n): q = deque([0]) res = [] while q: hp = [] for _ in range(len(q)): cur = q.popleft() heappush(hp, arr[cur]) for i in [2 * cur + 1, 2 * cur + 2]: if i < len(arr): q.append(i) t = [] while hp: t.append(heappop(hp)) res.append(t) return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): (res, start, end, len1) = ([], 0, 1, 1) while start < n: res.append(sorted(arr[start:end])) len1 *= 2 start = end end = start + len1 return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): i = 1 ans = [] while len(arr): ans.append(sorted(arr[:i])) arr = arr[i:] i <<= 1 return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): i = 0 k = 1 res = [] while i < n: temp = arr[i:i + k] temp.sort() res.append(temp) i += k k *= 2 return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): _list = [] a = 1 curr = 0 while curr < n: _list.append(sorted(arr[curr:curr + a])) curr += a a *= 2 return _list

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): if n == 1: return [[arr[0]]] else: l = [[arr[0]]] i = 1 c = 1 while i < n: size = 2 ** c if i + size < n: a = arr[i:i + size] a.sort() l.append(a) i += size c += 1 else: a = arr[i:] a.sort() l.append(a) break return l

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): start = 0 i = 0 increment = 2 ** i list1 = [] while start < n: list1.append(sorted(arr[start:start + increment])) start += increment i += 1 increment = 2 ** i return list1

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): res = [] i = 0 count = 0 level = 0 while True: count = 2 ** level t = [] while count != 0 and i < n: t.append(arr[i]) i += 1 count -= 1 res.append(sorted(t)) if i >= n: break level += 1 return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): res = [] l = 0 i = 0 while True: count = int(2 ** l) tmp = [] while count != 0 and i < n: tmp.append(arr[i]) i += 1 count -= 1 res.append(sorted(tmp)) if i >= n: break l += 1 return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): i = 0 a = [] while 2 ** i <= n: a.append(sorted(arr[:2 ** i])) arr[:] = arr[2 ** i:] i = i + 1 return a

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): ans = [] i = 0 level = 0 while True: count = int(2 ** level) tmp = [] while count != 0 and i < n: tmp.append(arr[i]) i += 1 count -= 1 ans.append(sorted(tmp)) if i >= n: break level += 1 return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): from math import exp level = 0 prevLevelEnd = 0 out = [] while prevLevelEnd < n: nAtLevel = pow(2, level) out.append(list(sorted(arr[prevLevelEnd:prevLevelEnd + nAtLevel]))) prevLevelEnd += nAtLevel level += 1 return out

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): re = [] level = 0 i = 0 while i < n: ans = [] if level == 0: re.append([arr[i]]) i += 1 level += 1 else: size = 2 ** level if i + size < n: ans.extend(arr[i:i + size]) ans.sort() re.append(ans) i += size level += 1 else: ans.extend(arr[i:]) ans.sort() re.append(ans) break return re

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

import heapq class Solution: def binTreeSortedLevels(self, arr, n): lst = [] x = 0 y = 1 while True: ll = [] for j in range(x, min(x + y, n)): ll.append(arr[j]) lst.append(sorted(ll)) x = x + y y = 2 * y if x >= n: break return lst

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, x, n): res = [] i = 0 j = 1 while i < n: res.append(sorted(x[i:i + j])) i = i + j j = j * 2 return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): ans = [] si = 0 k = 0 while si < n: size = 2 ** k k += 1 if si + size >= n: tans = arr[si:] else: tans = arr[si:si + size] tans.sort() ans.append(tans) si += size return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): lst = [] level = 0 i = 0 while i < n: l = [] if level == 0: l.append(arr[i]) lst.append(l) i = i + 1 level = level + 1 else: size = 2 ** level if i + size < n: l.extend(arr[i:i + size]) l.sort() lst.append(l) i = i + size level = level + 1 else: l.extend(arr[i:]) l.sort() lst.append(l) break return lst

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): a = [[arr[0]]] i = 1 while i < n: b = arr[i:2 * i + 1] b.sort() a.append(b) i = 2 * i + 1 return a

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): c = 0 i = 0 ans = [] while c < n: l = [] x = pow(2, i) while x > 0 and c < n: l.append(arr[c]) c += 1 x -= 1 i += 1 l.sort() ans.append(l) return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

from collections import deque class Solution: def binTreeSortedLevels(self, arr, n): lqc = deque() lqc.append(0) lqn = deque() lsrl = deque() rslt = deque() while len(lqc) > 0: idx = lqc.popleft() lsrl.append(arr[idx]) if 2 * idx + 1 < n: lqn.append(2 * idx + 1) if 2 * idx + 2 < n: lqn.append(2 * idx + 2) if len(lqc) == 0: lqc = lqn.copy() lqn = deque() lsrl = list(lsrl) lsrl.sort() rslt.append(lsrl) lsrl = deque() return rslt

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): mm = 0 x = 0 b = [] if n == 1: return [arr] exit() while x < n: e = 2 ** mm t = [] for k in range(e): if x < n: t.append(arr[x]) x = x + 1 t.sort() mm = mm + 1 b.append(t) return b

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): i = 0 c = 0 b = [] while i < n: e = 2 ** c k = [] for j in range(e): if i < n: k.append(arr[i]) i += 1 k.sort() c += 1 b.append(k) return b t = int(input()) for tc in range(t): n = int(input()) arr = list(map(int, input().split())) ob = Solution() res = ob.binTreeSortedLevels(arr, n) for i in range(len(res)): for j in range(len(res[i])): print(res[i][j], end=' ') print()

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

import heapq class Solution: def binTreeSortedLevels(self, arr, n): if len(arr) == 1: temp = [] temp.append([arr[0]]) return temp if len(arr) == 2: temp = [] temp.append([arr[0]]) temp.append([arr[1]]) return temp q1 = [] index = 0 ans = [] q1.append([arr[0], 0]) q2 = [] flag = 1 while True: temp = [] if flag == 1: heapq.heapify(q1) while q1: p = heapq.heappop(q1) temp.append(p[0]) if p[1] * 2 + 1 < n: q2.append([arr[p[1] * 2 + 1], p[1] * 2 + 1]) if p[1] * 2 + 2 < n: q2.append([arr[p[1] * 2 + 2], p[1] * 2 + 2]) flag = 0 ans.append(temp) if len(q2) == 0: break else: heapq.heapify(q2) while q2: p = heapq.heappop(q2) temp.append(p[0]) if p[1] * 2 + 1 < n: q1.append([arr[p[1] * 2 + 1], p[1] * 2 + 1]) if p[1] * 2 + 2 < n: q1.append([arr[p[1] * 2 + 2], p[1] * 2 + 2]) ans.append(temp) flag = 1 if len(q1) == 0: break return ans t = int(input()) for tc in range(t): n = int(input()) arr = list(map(int, input().split())) ob = Solution() res = ob.binTreeSortedLevels(arr, n) for i in range(len(res)): for j in range(len(res[i])): print(res[i][j], end=' ') print()

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): a = [] i = p = 0 while i < len(arr): a.append(sorted(arr[i:i + 2 ** p])) i += 2 ** p p += 1 return a t = int(input()) for tc in range(t): n = int(input()) arr = list(map(int, input().split())) ob = Solution() res = ob.binTreeSortedLevels(arr, n) for i in range(len(res)): for j in range(len(res[i])): print(res[i][j], end=' ') print()

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

from collections import defaultdict import queue class Solution: def binTreeSortedLevels(self, arr, n): q = queue.deque() dic = defaultdict(list) q.append([arr[0], 0, 0]) while q: ele = q.popleft() dic[ele[1]].append(ele[0]) val = 2 * ele[2] if val + 1 < n: q.append([arr[val + 1], ele[1] + 1, val + 1]) if val + 2 < n: q.append([arr[val + 2], ele[1] + 1, val + 2]) for i in dic: dic[i].sort() return dic

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): l = 0 el = 1 r = l + el ans = [] while r <= len(arr): brr = arr[l:r] brr.sort() ans.append(brr) el *= 2 if r < len(arr): l = r if l + el <= len(arr): r = l + el else: r = len(arr) elif r == len(arr): break return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): if not arr: return res = [[arr[0]]] level = 1 i = 1 l = len(arr) while i < l: tmp = [] j = 0 while i < l and j < 2 ** level: tmp.append(arr[i]) i += 1 j += 1 level += 1 tmp.sort() res.append(tmp) return res

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): ind = 0 ans = [] q = [arr[ind]] while q: b = sorted(q) ans.append(b) nn = len(q) for i in range(nn): p = q.pop(0) if ind + 1 < n: q.append(arr[ind + 1]) if ind + 2 < n: q.append(arr[ind + 2]) ind += 2 return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): lvl = -1 ans = [] while len(arr) > 0: lvl += 1 s = pow(2, lvl) if s > len(arr): s = len(arr) arr[0:s].sort() ans.append([]) while s > 0: ans[lvl].append(arr.pop(0)) s -= 1 for i in range(lvl + 1): ans[i].sort() return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): j = 0 l = [] while j < n: i = 2 * j + 1 l1 = arr[j:i] l1.sort() l.append(l1) j = i return l

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

class Solution: def binTreeSortedLevels(self, arr, n): c = 0 p = 0 l = [] while p < n: s = 2 ** c k = arr[p:p + s] c = c + 1 p = p + s k.sort() l.append(k) return l

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco

from collections import deque class Solution: def binTreeSortedLevels(self, arr, n): (i, j, k) = (0, 0, 0) ans = [] while j < n: st = [] i = 2 ** k while i > 0 and j < n: st.append(arr[j]) j += 1 i -= 1 ans.append(sorted(st)) k += 1 return ans

python

Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order. Example 1: Input: N = 7 arr[] = {7 6 5 4 3 2 1} Output: 7 5 6 1 2 3 4 Explanation: The formed Binary Tree is: 7 / \ 6 5 / \ / \ 4 3 2 1 Example 2: Input: N = 6 arr[] = {5 6 4 9 2 1} Output: 5 4 6 1 2 9 Explanation: The formed Binary Tree is: 5 / \ 6 4 / \ / 9 2 1 Your Task: You don't need to read input or print anything. Your task is to complete the function binTreeSortedLevels() which takes the array arr[] and its size N as inputs and returns a 2D array where the i-th array denotes the nodes of the i-th level in sorted order. Expected Time Complexity: O(NlogN). Expected Auxiliary Space: O(N). Constraints: 1 <= N <= 10^{4}

taco