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https://leetcode.com/problems/two-sum/discuss/2361743/Python-Simple-Solution-oror-O(n)-Time-oror-O(n)-Space
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: d = {} for i, j in enumerate(nums): r = target - j if r in d: return [d[r], i] d[j] = i # An Upvote will be encouraging
two-sum
Python Simple Solution || O(n) Time || O(n) Space
rajkumarerrakutti
288
21,600
two sum
1
0.491
Easy
0
https://leetcode.com/problems/two-sum/discuss/1378197/Simple-oror-100-faster-oror-5-Lines-code-oror-Well-Explained
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: store = dict() for i in range(len(nums)): sec = target - nums[i] if sec not in store: store[nums[i]]=i else: return [store[sec],i]
two-sum
🐍 Simple || 100% faster || 5 Lines code || Well-Explained 📌📌
abhi9Rai
47
6,800
two sum
1
0.491
Easy
1
https://leetcode.com/problems/two-sum/discuss/1378197/Simple-oror-100-faster-oror-5-Lines-code-oror-Well-Explained
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: tmp = sorted(nums) n, i, j = len(nums), 0, (n-1) while True: s = tmp[i]+tmp[j] if s>target: j-=1 elif s<target: i+=1 else: break return [nums.index(tmp[i]),n-(nums[::-1].index(tmp[j]))-1]
two-sum
🐍 Simple || 100% faster || 5 Lines code || Well-Explained 📌📌
abhi9Rai
47
6,800
two sum
1
0.491
Easy
2
https://leetcode.com/problems/two-sum/discuss/2045849/1LINER-O(n)timeBEATS-99.97-MEMORYSPEED-0ms-MAY-2022
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: seen = {} for i, value in enumerate(nums): #1 remaining = target - nums[i] #2 if remaining in seen: #3 return [i, seen[remaining]] #4 else: seen[value] = i #5
two-sum
[1LINER] O(n)time/BEATS 99.97% MEMORY/SPEED 0ms MAY 2022
cucerdariancatalin
21
4,800
two sum
1
0.491
Easy
3
https://leetcode.com/problems/two-sum/discuss/1680976/Python-Simple-O(n)-time-solution-with-explanation-and-Big-O-analysis
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: prevTable = {} for i,currVal in enumerate(nums): complement = target - currVal if complement in prevTable: return [prevTable[complement],i] prevTable[currVal] = i
two-sum
Python Simple O(n) time solution, with explanation and Big O analysis
kenanR
20
2,400
two sum
1
0.491
Easy
4
https://leetcode.com/problems/two-sum/discuss/1855612/Python3-direct-solution-faster-than-85.68-online-submissions-greater-O(n)
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: map = {} for i,n in enumerate(nums): diff = target - n if diff in map: return [map[diff],i] map[n] = i
two-sum
Python3 direct solution faster than 85.68% online submissions -> O(n)
MaxKingson
13
1,100
two sum
1
0.491
Easy
5
https://leetcode.com/problems/two-sum/discuss/2806290/Python-4-Line-simple-solution
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: dict_nums = {} for i, v in enumerate(nums): if v in dict_nums: return [i, dict_nums[v]] dict_nums[target - v] = i
two-sum
😎 Python 4-Line simple solution
Pragadeeshwaran_Pasupathi
7
2,200
two sum
1
0.491
Easy
6
https://leetcode.com/problems/two-sum/discuss/2391581/python3-O(n)-and-O(n2)-both-made-easy.-simple
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)-1): for j in range(len(nums)): if j!=i: if (target == nums[i]+nums[j]): return i,j
two-sum
python3 O(n) and O(n2) both made easy. simple
md-thayyib
7
734
two sum
1
0.491
Easy
7
https://leetcode.com/problems/two-sum/discuss/2391581/python3-O(n)-and-O(n2)-both-made-easy.-simple
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: lookup = {} for position, number in enumerate(nums): if target - number in lookup: return lookup[target-number],position else: lookup[number]=position
two-sum
python3 O(n) and O(n2) both made easy. simple
md-thayyib
7
734
two sum
1
0.491
Easy
8
https://leetcode.com/problems/two-sum/discuss/1744681/100-Python3-or-Faster-Solution-or-Easiest
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: checked={} for index , item in enumerate(nums): remaining = target - nums[index] if remaining in checked: return [index, checked[remaining]] checked[item] = index
two-sum
✔ 100% Python3 | Faster Solution | Easiest
Anilchouhan181
7
933
two sum
1
0.491
Easy
9
https://leetcode.com/problems/two-sum/discuss/1634584/Python-Easy-Solution-or-Simple-Approach
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: res = {} for idx, val in enumerate(nums): remn = target-val if remn in res: return [res[remn], idx] res[val] = idx
two-sum
Python Easy Solution | Simple Approach ✔
leet_satyam
7
1,400
two sum
1
0.491
Easy
10
https://leetcode.com/problems/two-sum/discuss/1503197/Move-from-O(N2)-to-O(N)-oror-Thought-Process-Explained
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: #Brute Force way of thinking #Generate all subarrays and as soon as we find the condition getting fulfilled, append it to our answer #Note - as per question, only 1 valid answer exists, so repetition will not be an issue #If at all there was repetition, it can be avoided by traversing as below #O(N^2) Time -- Accepted Solution n = len(nums) answer = list() for i in range(n): for j in range(i+1,n): if (nums[i] + nums[j] == target): answer.append(i) answer.append(j) return answer #Let us try to optimize #NOTE - We CANNOT SORT THIS array as we have to retrieve the indices and sorting it #would change the original indices at which elements are present #If the given array would have already been sorted, this would have worked completely fine #if we fix one of the numbers as arr[i], so other would be arr[j] = (target - arr[i]) #now, we simply have to see if this arr[j] exists in the given array or not #O(N) Time --- Accepted Solution (TAKEN HELP FROM DISCUSSION SECTION) #NOTE - Looking up an element in a dictionary is O(1) and not O(N) n = len(nums) seen = {} #empty dictionary for index, value in enumerate(nums): remaining = target - value #here, value is nums[i] #we are looking for this remaining, which is nums[j] #we have to find if remaining or nums[j] is present in the given array or not #this equation comes from the fact that as per question : #nums[i] + nums[j] = target if remaining in seen: #if nums[j] is present in dictionary #just return index of nums[i] and nums[j] return [index, seen[remaining]] #index represents index of nums[i] and seen[remaining] will #hold index of nums[j] else: #otherwise, add the value corresponding to its index in the dictionary #so, if it searched again, we can directly retrieve its index seen[value] = index
two-sum
Move from O(N^2) to O(N) || Thought Process Explained
aarushsharmaa
7
890
two sum
1
0.491
Easy
11
https://leetcode.com/problems/two-sum/discuss/2287919/Easy-python3-using-enumerate(updated)
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: prevMap = {} for i, n in enumerate(nums): diff = target - n if diff in prevMap: return [prevMap[diff], i] prevMap[n] = i
two-sum
Easy python3 using enumerate(updated)
__Simamina__
6
513
two sum
1
0.491
Easy
12
https://leetcode.com/problems/two-sum/discuss/1812148/PythonJava-Hash-Map-of-Inverses-or-Beats-99
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: inverses = {} for i, num in enumerate(nums): if num in inverses: return [inverses[num], i] inverses[target-num] = i
two-sum
[Python/Java] Hash Map of Inverses | Beats 99%
hari19041
6
650
two sum
1
0.491
Easy
13
https://leetcode.com/problems/two-sum/discuss/1400832/Python-or-Easy-or-O(n)
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: #two pointer technique with sorting -> time -> O(nlogn) #hashing technique -> time -> O(n) di = {} for i in range(len(nums)): if target-nums[i] in di: return [di[target-nums[i]], i] else: di[nums[i]] = i #no need we will find solution in loop itself acc. to the question
two-sum
Python | Easy | O(n)
sathwickreddy
6
1,600
two sum
1
0.491
Easy
14
https://leetcode.com/problems/two-sum/discuss/1837631/Accepted-Python3-solution-with-2-lines
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in nums: if target - i in nums[nums.index(i) + 1:]: return [nums.index(i), nums[nums.index(i) + 1:].index(target - i) + nums.index(i) + 1]
two-sum
Accepted Python3 solution with 2 lines
ElizaZoldyck
5
608
two sum
1
0.491
Easy
15
https://leetcode.com/problems/two-sum/discuss/1769992/Python-solution-using-hash-map-with-O(n)-complexity
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: # construnct a empty hash map val_index_map = {} # hasmpa { value: index} # iterate through given array with count, I used enumerate for count you guys can also add its mannually for count, num in enumerate(nums): # first i check the diff btw target and current value req_diff = target - num # if required difrence is exist in hashmap, then we are done if req_diff in val_index_map: # so current_value + required_diff = Traget # mission accomplished return [val_index_map.get(req_diff), count] else: # if required diff is not present in the hasmap then we add current number as a value and its index as a key val_index_map[num] = count return
two-sum
Python solution using hash map with O(n) complexity
rajat4665
5
929
two sum
1
0.491
Easy
16
https://leetcode.com/problems/two-sum/discuss/2737044/Python-Hash-table-O(n)-or-Full-explanation
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: """Take each number and check if its complement has been seen before. If not, add it to the list of known complements along with its index. Args: nums (List[int]): Input array of integers target (int): Target integer Returns: List[int]: Indices of the two integers that sum to target """ # Initialise hash map to store known integers complements = {} # Iterate through the list for i in range(len(nums)): # Check if the current number's complement has been seen before complement = target - nums[i] if complement in complements: return [complements[complement], i] # Add the current number to the list of known complements complements[nums[i]] = i
two-sum
🥇 [Python] Hash table - O(n) | Full explanation ✨
LloydTao
4
613
two sum
1
0.491
Easy
17
https://leetcode.com/problems/two-sum/discuss/2737044/Python-Hash-table-O(n)-or-Full-explanation
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: """Take each pair of numbers and see if they add up to the target. Args: nums (List[int]): Input array of integers target (int): Target integer Returns: List[int]: Indices of the two integers that sum to target """ # Get length of input array n = len(nums) # Iterate over all pairs (i,j) for i in range(n): for j in range(i + 1, n): # Check if this pair equals the target if nums[i] + nums[j] == target: return [i, j]
two-sum
🥇 [Python] Hash table - O(n) | Full explanation ✨
LloydTao
4
613
two sum
1
0.491
Easy
18
https://leetcode.com/problems/two-sum/discuss/2603308/easy-python-solution
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)) : num1 = nums[i] if (target - num1) in nums[i+1:] : return [i, (nums[i+1:].index((target - num1)))+i+1]
two-sum
easy python solution
sghorai
4
806
two sum
1
0.491
Easy
19
https://leetcode.com/problems/two-sum/discuss/2570856/SIMPLE-PYTHON3-SOLUTION
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): try: if i != nums.index(target - nums[i]): return [i, nums.index(target - nums[i])] except ValueError: continue
two-sum
✅✔ SIMPLE PYTHON3 SOLUTION ✅✔
rajukommula
4
339
two sum
1
0.491
Easy
20
https://leetcode.com/problems/two-sum/discuss/2546192/Simple-dictionary-solution
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: pairs = {} for i, v in enumerate(nums): if target-v in pairs: return [pairs[target-v], i] else: pairs[v] = i
two-sum
📌 Simple dictionary solution
croatoan
4
226
two sum
1
0.491
Easy
21
https://leetcode.com/problems/two-sum/discuss/2542771/1.-Two-Sum
class Solution: def twoSum(self, nums: list[int], target: int) -> list[int]: s =defaultdict(int) for i, n in enumerate(nums): if target - n in s: return [i, s[target - n]] s[n] =i
two-sum
1. Two Sum
warrenruud
4
543
two sum
1
0.491
Easy
22
https://leetcode.com/problems/two-sum/discuss/2454524/Optimal-PYTHON-3-Solution
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: hashmap = {} # val : index for i, n in enumerate(nums): #if n<=target: diff = target-n if diff in hashmap: return [hashmap[diff],i] hashmap[n]=i return
two-sum
Optimal PYTHON 3 Solution
WhiteBeardPirate
4
473
two sum
1
0.491
Easy
23
https://leetcode.com/problems/two-sum/discuss/1691722/**-Python-code%3A-using-HashMap
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: checked={} for index , item in enumerate(nums): remaining = target - nums[index] if remaining in checked: return [index, checked[remaining]] checked[item] = index
two-sum
** Python code: using HashMap
Anilchouhan181
4
407
two sum
1
0.491
Easy
24
https://leetcode.com/problems/two-sum/discuss/444887/Python3C%2B%2B-hash-table
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: seen = dict() #hash table of value-index pair for i, num in enumerate(nums): x = target - num #residual if x in seen: return [seen[x], i] seen[num] = i return None
two-sum
[Python3/C++] hash table
ye15
4
774
two sum
1
0.491
Easy
25
https://leetcode.com/problems/two-sum/discuss/398550/Two-Solutions-in-Python-3-(Dictionary)-(-O(n)-)
class Solution: def twoSum(self, N: List[int], t: int) -> List[int]: D = {n:i for i,n in enumerate(N)} for i,n in enumerate(N): x = t - n if x in D and D[x] != i: return [i,D[x]]
two-sum
Two Solutions in Python 3 (Dictionary) ( O(n) )
junaidmansuri
4
1,800
two sum
1
0.491
Easy
26
https://leetcode.com/problems/two-sum/discuss/398550/Two-Solutions-in-Python-3-(Dictionary)-(-O(n)-)
class Solution: def twoSum(self, N: List[int], t: int) -> List[int]: D = {} for i,n in enumerate(N): if n not in D: D[n] = i x = t - n if x in D and D[x] != i: return [i,D[x]] - Junaid Mansuri (LeetCode ID)@hotmail.com
two-sum
Two Solutions in Python 3 (Dictionary) ( O(n) )
junaidmansuri
4
1,800
two sum
1
0.491
Easy
27
https://leetcode.com/problems/two-sum/discuss/384034/super-basic-question...
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]:
two-sum
super basic question...
jstarrk
4
815
two sum
1
0.491
Easy
28
https://leetcode.com/problems/two-sum/discuss/2762217/Python-Simple-Solution-oror-Easy-to-Understand
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)-1): for j in range(i+1,len(nums)): if nums[i]+nums[j]==target: return [i,j]
two-sum
🐍🐍Python Simple Solution || Easy to Understand✅✅
sourav638
3
43
two sum
1
0.491
Easy
29
https://leetcode.com/problems/two-sum/discuss/2724248/Python's-Simple-and-Easy-to-Understand-Solution-Using-Dictionary-or-99-Faster
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: # Creating Dictonary for Lookup lookup = {} for i in range(len(nums)): # if target-n in lookup return n index and current index if target-nums[i] in lookup: return [lookup[target-nums[i]], i] lookup[nums[i]] = i
two-sum
✔️ Python's Simple and Easy to Understand Solution Using Dictionary | 99% Faster 🔥
pniraj657
3
639
two sum
1
0.491
Easy
30
https://leetcode.com/problems/two-sum/discuss/2671221/Python-three-solutions
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): for j in range(i + 1, len(nums)): if nums[i] + nums[j] == target: return [i, j]
two-sum
✔️ Python three solutions
QuiShimo
3
460
two sum
1
0.491
Easy
31
https://leetcode.com/problems/two-sum/discuss/2671221/Python-three-solutions
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): find_num = target - nums[i] if find_num in nums and nums.index(find_num) != i: return [i, nums.index(find_num)]
two-sum
✔️ Python three solutions
QuiShimo
3
460
two sum
1
0.491
Easy
32
https://leetcode.com/problems/two-sum/discuss/2671221/Python-three-solutions
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: nums_dict = {} for i, j in enumerate(nums): find_number = target - nums[i] if find_number in nums_dict: return [nums_dict[find_number], i] nums_dict[j] = i
two-sum
✔️ Python three solutions
QuiShimo
3
460
two sum
1
0.491
Easy
33
https://leetcode.com/problems/two-sum/discuss/2269969/Python3-solution-using-enumerate
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: prevMap = {} for i, n in enumerate(nums): diff = target - n if diff in prevMap: return [prevMap[diff], i] prevMap[n] = i
two-sum
Python3 solution using enumerate
__Simamina__
3
451
two sum
1
0.491
Easy
34
https://leetcode.com/problems/two-sum/discuss/2073481/Python-easy-to-understand-and-read-or-sorting
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: arr = [] for i, num in enumerate(nums): arr.append([i, num]) arr.sort(key=lambda x:x[1]) ans = [] i, j = 0, len(arr)-1 while i < j: sums = arr[i][1] + arr[j][1] if sums == target: return [arr[i][0], arr[j][0]] if sums > target: j = j-1 else: i = i+1
two-sum
Python easy to understand and read | sorting
sanial2001
3
635
two sum
1
0.491
Easy
35
https://leetcode.com/problems/two-sum/discuss/1753486/brute-force
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): for j in range(i + 1, len(nums)): if nums[i] + nums[j] == target: return [i, j]
two-sum
brute force
ggeeoorrggee
3
238
two sum
1
0.491
Easy
36
https://leetcode.com/problems/two-sum/discuss/1653823/O(n)-Solution-using-dictionary-in-Python
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: # O(n) solution using HashMap or Dictionary dict_nums = {} for idx in range(0, len(nums)): curr_num = nums[idx] diff = target - curr_num if diff in dict_nums: return [idx, dict_nums[diff]] else: dict_nums[curr_num] = idx # # Brute Force - O(n2) # for i in range(0, len(nums)-1): # for j in range(i+1, len(nums)): # if nums[i] + nums[j] == target: # return [i, j] ```
two-sum
O(n) Solution using dictionary in Python
coddify
3
275
two sum
1
0.491
Easy
37
https://leetcode.com/problems/two-sum/discuss/1630535/Python-3-oror-brute-force
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): for j in range(i): if nums[i]+nums[j]==target: return [i,j]
two-sum
Python 3 || brute force
anon_python_coder
3
820
two sum
1
0.491
Easy
38
https://leetcode.com/problems/two-sum/discuss/1392824/Python3-96-faster-(52-ms)-Hash-Table-Solution
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: D = {} for i in range(len(nums)): temp = target-nums[i] if temp in D: return [i, D[temp]] else: D[nums[i]] = i
two-sum
[Python3] 96% faster (52 ms) - Hash Table Solution
Loqz
3
1,800
two sum
1
0.491
Easy
39
https://leetcode.com/problems/two-sum/discuss/1124970/Python3-Solution-using-Hashmap-by-dictionary
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: map = {} for i in range(len(nums)): map[nums[i]] = i for i in range(len(nums)): value = target - nums[i] if value in map and map[value] != i: return [i, map[value]]
two-sum
Python3 Solution using Hashmap by dictionary
vtthanh99
3
487
two sum
1
0.491
Easy
40
https://leetcode.com/problems/two-sum/discuss/967361/Python3-O(n)-with-explanation
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: seen = {} for i in range(len(nums)): key = target-nums[i] if key in seen: return [seen[key], i] seen[nums[i]] = i
two-sum
[Python3] O(n) with explanation
gdm
3
527
two sum
1
0.491
Easy
41
https://leetcode.com/problems/two-sum/discuss/2426335/Python-Solution
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: hashMap = {} for i in range(len(nums)): hashMap[nums[i]] = i for i in range(len(nums)): rem = target -nums[i] if rem in hashMap and hashMap[rem] != i: return [i,hashMap[rem]] #Upvote will be encouraging.
two-sum
Python Solution
anshsharma17
2
341
two sum
1
0.491
Easy
42
https://leetcode.com/problems/two-sum/discuss/2000886/Python-3-Solution-oror-Three-Approaches
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: # For storing answers in one hashmap also will check if any value exists in nums checker = {} for i, n in enumerate(nums): diff_value = target - n # If value already in answer checker if diff_value in checker: return [checker[diff_value],i] checker[n] = i
two-sum
✅ Python 3 Solution || Three Approaches
mitchell000
2
459
two sum
1
0.491
Easy
43
https://leetcode.com/problems/two-sum/discuss/2000886/Python-3-Solution-oror-Three-Approaches
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: # For storing answers in one set also will check if any value exists in nums checker = set() for i in range(len(nums)): # If value already in answer checker if nums[i] in checker: return [nums.index(target-nums[i]),i] else: checker.add(target-nums[i])
two-sum
✅ Python 3 Solution || Three Approaches
mitchell000
2
459
two sum
1
0.491
Easy
44
https://leetcode.com/problems/two-sum/discuss/2000886/Python-3-Solution-oror-Three-Approaches
class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: left = 0 right = len(nums) while left < right: if target-nums[left] in nums and nums.index(target-nums[left]) != left: return [left,nums.index(target-nums[left])] else: left+=1
two-sum
✅ Python 3 Solution || Three Approaches
mitchell000
2
459
two sum
1
0.491
Easy
45
https://leetcode.com/problems/add-two-numbers/discuss/1835217/Python3-DUMMY-CARRY-(-**-)-Explained
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: res = dummy = ListNode() carry = 0 while l1 or l2: v1, v2 = 0, 0 if l1: v1, l1 = l1.val, l1.next if l2: v2, l2 = l2.val, l2.next val = carry + v1 + v2 res.next = ListNode(val%10) res, carry = res.next, val//10 if carry: res.next = ListNode(carry) return dummy.next
add-two-numbers
✔️ [Python3] DUMMY CARRY ( •⌄• ू )✧, Explained
artod
44
7,100
add two numbers
2
0.398
Medium
46
https://leetcode.com/problems/add-two-numbers/discuss/452442/Python-3%3A-recursion
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: _ = l1.val + l2.val digit, tenth = _ % 10, _ // 10 answer = ListNode(digit) if any((l1.next, l2.next, tenth)): l1 = l1.next if l1.next else ListNode(0) l2 = l2.next if l2.next else ListNode(0) l1.val += tenth answer.next = self.addTwoNumbers(l1, l2) return answer
add-two-numbers
Python 3: recursion
deleted_user
39
4,200
add two numbers
2
0.398
Medium
47
https://leetcode.com/problems/add-two-numbers/discuss/486839/Python-Simple-Solution-8-Liner
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: sumval = 0 root = curr = ListNode(0) while l1 or l2 or sumval: if l1: sumval += l1.val; l1 = l1.next if l2: sumval += l2.val; l2 = l2.next curr.next = curr = ListNode(sumval % 10) sumval //= 10 return root.next
add-two-numbers
Python - Simple Solution - 8 Liner
mmbhatk
18
4,600
add two numbers
2
0.398
Medium
48
https://leetcode.com/problems/add-two-numbers/discuss/2024907/Simple-Python3-Solution
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: dummy = ListNode() cur = dummy carry = 0 while l1 or l2 or carry: v1 = l1.val if l1 else 0 v2 = l2.val if l2 else 0 # new digit val val = v1 + v2 + carry carry = val // 10 val = val % 10 cur.next = ListNode(val) # as in one place we have to put a single digit # update pointers cur = cur.next l1 = l1.next if l1 else None l2 = l2.next if l2 else None return dummy.next # Time: O(n + m) # Space: O(n + m)
add-two-numbers
Simple Python3 Solution✨
samirpaul1
11
1,000
add two numbers
2
0.398
Medium
49
https://leetcode.com/problems/add-two-numbers/discuss/1353295/Python-Runtime-60ms
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: start = curr = ListNode(0) carry = 0 while(l1 or l2 or carry): x = l1.val if l1 else 0 y = l2.val if l2 else 0 carry, val = divmod(x + y + carry, 10) curr.next = ListNode(val) curr = curr.next l1 = l1.next if l1 else None l2 = l2.next if l2 else None return start.next
add-two-numbers
[Python] Runtime 60ms
yadvendra
11
1,200
add two numbers
2
0.398
Medium
50
https://leetcode.com/problems/add-two-numbers/discuss/2383456/Fastest-Solution-Explained0ms100-O(n)time-complexity-O(n)space-complexity
class Solution: def addTwoNumbers(self, l1, l2): list1 = make_list_from_ListNode(l1) list2 = make_list_from_ListNode(l2) ### RIGHT PAD WITH ZEROES len_list1 = len(list1) len_list2 = len(list2) if len_list1 > len_list2: pad = len_list1 - len_list2 list2 = list2 + [0,] * pad elif len_list2 > len_list1: pad = len_list2 - len_list1 list1 = list1 + [0,] * pad ### DO THE MATH d = 0 the_sum = list() for x,y in zip(list1, list2): d, m = divmod(x + y + d, 10) the_sum.append(m) if d != 0: the_sum.append(d) return make_ListNode_from_list(the_sum)
add-two-numbers
[Fastest Solution Explained][0ms][100%] O(n)time complexity O(n)space complexity
cucerdariancatalin
10
1,700
add two numbers
2
0.398
Medium
51
https://leetcode.com/problems/add-two-numbers/discuss/1636945/Python3-O(1)-Space-or-1-pass-or-Optimal-or-Beats-100
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: carry, head = 0, l1 while l1 or l2: if not l1.next and l2: l1.next, l2.next = l2.next, l1.next val1 = l1.val if l1 else 0 val2 = l2.val if l2 else 0 carry, l1.val = divmod(val1 + val2 + carry, 10) prev = l1 if l1: l1 = l1.next if l2: l2 = l2.next if carry: prev.next = ListNode(carry) return head
add-two-numbers
[Python3] O(1) Space | 1 pass | Optimal | Beats 100%
PatrickOweijane
9
814
add two numbers
2
0.398
Medium
52
https://leetcode.com/problems/add-two-numbers/discuss/1784472/Python-3-greater-Simple-solution-that-was-asked-in-real-interview
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: carryOver = 0 result = ListNode(-1) resultTail = result while l1 or l2: total = 0 if l1: total += l1.val l1 = l1.next if l2: total += l2.val l2 = l2.next total += carryOver carryOver, remainder = divmod(total, 10) resultTail.next = ListNode(remainder) resultTail = resultTail.next if carryOver > 0: resultTail.next = ListNode(carryOver) return result.next
add-two-numbers
Python 3 -> Simple solution that was asked in real interview
mybuddy29
8
637
add two numbers
2
0.398
Medium
53
https://leetcode.com/problems/add-two-numbers/discuss/468242/Easy-to-follow-Python3-solution-(faster-than-96-and-memory-usage-less-than-100)
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: r = None p = None c = 0 while l1 or l2 or c != 0: v = (l1.val if l1 else 0) + (l2.val if l2 else 0) + c c = v // 10 n = ListNode(v % 10) if r is None: r = n if p is not None: p.next = n p = n l1 = l1.next if l1 else None l2 = l2.next if l2 else None return r
add-two-numbers
Easy-to-follow Python3 solution (faster than 96% & memory usage less than 100%)
gizmoy
7
1,400
add two numbers
2
0.398
Medium
54
https://leetcode.com/problems/add-two-numbers/discuss/1836636/Python-Easy-Solution-or-Explained
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: head = sm = ListNode() # head: we get the start to return, sm: to iterate and link the new sum nodes. carry = 0 # carry starts with zero while l1 or l2: # we are using or, because list sizes might not be equal. if l1: carry, l1 = carry + l1.val, l1.next # if l1 is not null, add it to carry and update l1 to its next. if l2: carry, l2 = carry + l2.val, l2.next # if l2 is not null, add it to carry and update l2 to its next. sm.next = sm = ListNode(val = carry % 10) # make a new node with carry % 10(because carry can be greater than 9, ie two digits) and link the sm.next to it and now this new node is our sm carry //= 10 # if carry is greater than 9, means we have to add it in next iteration if carry: sm.next = ListNode(val = carry) # what if there is non zero carry left to be added. return head.next # return head.next bcoz head is pointing to dummy node.
add-two-numbers
✅ Python Easy Solution | Explained
dhananjay79
5
397
add two numbers
2
0.398
Medium
55
https://leetcode.com/problems/add-two-numbers/discuss/2308917/Python3-using-two-pointers
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: dummy = ListNode() cur = dummy carry = 0 while l1 or l2 or carry: v1 = l1.val if l1 else 0 v2 = l2.val if l2 else 0 val = v1 + v2 + carry carry = val // 10 val = val % 10 cur.next = ListNode(val) cur = cur.next l1 = l1.next if l1 else None l2 = l2.next if l2 else None return dummy.next
add-two-numbers
Python3 using two pointers
__Simamina__
4
249
add two numbers
2
0.398
Medium
56
https://leetcode.com/problems/add-two-numbers/discuss/1369722/python-3-solution-easyy
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: carry=0 h=ListNode() head=h while l1 or l2 or carry: result=0 if l1: result+=l1.val l1=l1.next if l2: result+=l2.val l2=l2.next if carry: result+=carry val=result%10 carry=result//10 head.next=ListNode(val) head=head.next return h.next
add-two-numbers
python 3 solution easyy
minato_namikaze
4
535
add two numbers
2
0.398
Medium
57
https://leetcode.com/problems/add-two-numbers/discuss/1515075/Python-solution-explained-(linked-list-greater-list-and-slice)
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: list_l1 = [] list_l1_int = '' list_l2 = [] list_l2_int = '' # convert linked lists to lists while l1: list_l1.append(l1.val) l1 = l1.next while l2: list_l2.append(l2.val) l2 = l2.next # iterate thru lists in reverse and recreate int from list for x in list_l1[::-1]: list_l1_int += str(x) for x in list_l2[::-1]: list_l2_int += str(x) # sum list_sum = int(list_l1_int) + int(list_l2_int) # now create list node from the sum (use s, t to return linked list at right iteration) s = t = ListNode() # conv to str, iterate thru items in reverse, create linked list for x in str(list_sum)[::-1]: s.next = ListNode(x) s = s.next return t.next
add-two-numbers
Python solution explained (linked list -> list & slice)
jjluxton
3
964
add two numbers
2
0.398
Medium
58
https://leetcode.com/problems/add-two-numbers/discuss/1222978/python3-%3A-Easy-to-understand-and-straight-forward-approach
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: # base cases if l1==None: return l2 if l2==None: return l1 curr = ListNode(0) head = curr carry = 0 while l1 and l2: total = 0 total += carry + l1.val + l2.val curr.next = ListNode(total%10) carry = total // 10 curr = curr.next l1 = l1.next l2 = l2.next while (l1): total = 0 total += carry + l1.val curr.next = ListNode(total%10) carry = total // 10 curr = curr.next l1 = l1.next while (l2): total = 0 total += carry + l2.val curr.next = ListNode(total%10) carry = total // 10 curr = curr.next l2 = l2.next if carry: curr.next = ListNode(carry) return head.next
add-two-numbers
python3 : Easy to understand and straight forward approach
nandanabhishek
3
353
add two numbers
2
0.398
Medium
59
https://leetcode.com/problems/add-two-numbers/discuss/1081524/python3-easy-solution-with-detailed-comments-faster-than-96.55
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: # Keep a dummy head for your return sentinel = ListNode(0) curr = sentinel # Initializes two pointers pointing to two lists seperately node1, node2 = l1, l2 # Check if we need add '1' to next calculation carry = False while node1 or node2: # Calculate the current sum, if node is None then initializes to 0 i = node1.val if node1 else 0 y = node2.val if node2 else 0 total = i + y # Check carry and update current sum if carry: total += 1 carry = False # Modify total sum to one digit number and update carry if total >= 10: carry = True total -= 10 # Initializes a new node based on current sum and link to the dummy head to_add = ListNode(total) curr.next = to_add curr = to_add # Update node1 and node2 for next calculation node1 = node1.next if node1 else None node2 = node2.next if node2 else None # Check carry at the end of calculation if carry: extraDigit = ListNode(1) curr.next = extraDigit return sentinel.next
add-two-numbers
python3 easy solution with detailed comments, faster than 96.55%
IssacZ
3
554
add two numbers
2
0.398
Medium
60
https://leetcode.com/problems/add-two-numbers/discuss/972330/Python3-solution
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: first_num=[] second_num=[] while l1 is not None: first_num.append(l1.val) l1=l1.next while l2 is not None: second_num.append(l2.val) l2=l2.next first_num.reverse() second_num.reverse() res1=int(''.join(map(str,first_num))) res2=int(''.join(map(str,second_num))) res3=str(res1+res2) l3=None for i in res3: data=int(i) newnode=ListNode(data) if not l3: l3=newnode else: newnode.next=l3 l3=newnode return l3
add-two-numbers
Python3 solution
samarthnehe
3
453
add two numbers
2
0.398
Medium
61
https://leetcode.com/problems/add-two-numbers/discuss/2214565/Python3-or-Easy-to-Understand-or-Simple
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: result = ListNode() result_tail = result carry = 0 while l1 or l2 or carry: val1 = l1.val if l1 else 0 val2 = l2.val if l2 else 0 out = (val1 + val2 + carry) % 10 carry = (val1 + val2 + carry) // 10 result_tail.next = ListNode(out) result_tail = result_tail.next l1 = l1.next if l1 else None l2 = l2.next if l2 else None return result.next
add-two-numbers
✅Python3 | Easy to Understand | Simple
thesauravs
2
120
add two numbers
2
0.398
Medium
62
https://leetcode.com/problems/add-two-numbers/discuss/2127776/EASY-Python-Solution-O(n)
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: s1, s2 = '', '' while l1: s1 += str(l1.val) l1 = l1.next while l2: s2 += str(l2.val) l2 = l2.next s1 = int(s1[::-1]) s2 = int(s2[::-1]) total = str(s1 + s2) head = None for c in total: n = ListNode(c) n.next = head head = n return head
add-two-numbers
EASY Python Solution O(n)
anselchacko
2
406
add two numbers
2
0.398
Medium
63
https://leetcode.com/problems/add-two-numbers/discuss/2079491/Python-Very-Intuitive-Solution
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: int_1 = int_2 = 0 i = 0 while l1 or l2: if l1: int_1 += l1.val * (10 ** i) l1 = l1.next if l2: int_2 += l2.val * (10 ** i) l2 = l2.next i += 1 int_sum = int_1 + int_2 for i, char in enumerate(list(str(int_sum))): if i == 0: node = ListNode(int(char), None) else: node = ListNode(int(char), node) return node
add-two-numbers
Python Very Intuitive Solution
codeee5141
2
378
add two numbers
2
0.398
Medium
64
https://leetcode.com/problems/add-two-numbers/discuss/1837667/Accepted-Python3-solution-DUMMY-DUMMY
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: if l1 == None and l2 == None: return ListNode(0) a = b = i = 0 while l1 != None: a += l1.val * (10 ** i) i += 1 l1 = l1.next i = 0 while l2 != None: b += l2.val * (10 ** i) i += 1 l2 = l2.next ans = a + b x = [int(l) for l in str(ans)][::-1] ansll = dummy = ListNode() for i in x[:-1]: ansll.val = i ansll.next = ListNode() ansll = ansll.next ansll.val = x[-1] return dummy
add-two-numbers
Accepted Python3 solution DUMMY DUMMY
ElizaZoldyck
2
54
add two numbers
2
0.398
Medium
65
https://leetcode.com/problems/add-two-numbers/discuss/1811728/PythonJava-Simple-Addition-Algorithm-or-Digit-by-Digit
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode: temp = ListNode(0) ptr1 = l1; ptr2 = l2 curr = temp; carry = 0 while ptr1 or ptr2: x = ptr1.val if ptr1 else 0 y = ptr2.val if ptr2 else 0 s = x+y+carry carry = s//10 curr.next = ListNode(s%10) curr = curr.next if ptr1: ptr1 = ptr1.next if ptr2: ptr2 = ptr2.next if carry: curr.next = ListNode(carry) return temp.next
add-two-numbers
[Python/Java] Simple Addition Algorithm | Digit by Digit
hari19041
2
537
add two numbers
2
0.398
Medium
66
https://leetcode.com/problems/add-two-numbers/discuss/1684721/Python-O(n)-and-in-place-beat-96
class Solution: def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ # O(n) and O(1) head = l1 carry, l1.val = divmod(l1.val + l2.val, 10) while l1.next and l2.next: l1 = l1.next l2 = l2.next carry, l1.val = divmod(l1.val + l2.val + carry, 10) if l2.next: l1.next = l2.next while l1.next: l1 = l1.next carry, l1.val = divmod(l1.val + carry, 10) if carry: l1.next = ListNode(carry) return head
add-two-numbers
Python, O(n) and in-place, beat 96%
yaok09
2
546
add two numbers
2
0.398
Medium
67
https://leetcode.com/problems/add-two-numbers/discuss/1409756/python-and-c%2B%2B-solution-(O(n)-most-efficient-solution-in-both-space-and-time-)
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: ans=ListNode() current=ans carry=0 while l1 and l2 : add=l1.val+l2.val+carry val=add%10 carry=add//10 current.next=ListNode(val) current=current.next l1=l1.next l2=l2.next current.next=l1 or l2 while carry>0: if current.next : current=current.next else: current.next=ListNode(carry) carry=0 add=carry+current.val val=add%10 carry=add//10 current.val=val return ans.next
add-two-numbers
python and c++ solution ,(O(n) most efficient solution in both space and time )
sagarhparmar12345
2
270
add two numbers
2
0.398
Medium
68
https://leetcode.com/problems/add-two-numbers/discuss/1401366/Python-easy-or-O(n)-or-Without-extra-space
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: h1, h2, prev = l1, l2, None carry = 0 while h1 != None and h2 != None: sm = h1.val+h2.val+carry carry = sm//10 h1.val = sm%10 prev = h1 h1 = h1.next h2 = h2.next if h2!=None: h1 = h2 #since we are overriding the first linked list, so our priority is on head1 if prev!=None: #if case will handle this edge case l1 = [] l2 = [1,2,3] prev.next = h2 while h1!=None: sm = h1.val+carry carry = sm//10 h1.val = sm%10 prev = h1 h1 = h1.next if carry: prev.next = ListNode(carry) return l1
add-two-numbers
Python easy | O(n) | Without extra space
sathwickreddy
2
909
add two numbers
2
0.398
Medium
69
https://leetcode.com/problems/add-two-numbers/discuss/262494/Python-faster-than-100-and-easy-to-read
class Solution: def addTwoNumbers(self, l1: ListNode, l2: ListNode, over=0) -> ListNode: if l1 is None and l2 is None: if over > 0: return ListNode(over) return None num = over next1 = None next2 = None if not l1 is None: num += l1.val next1 = l1.next if not l2 is None: num += l2.val next2 = l2.next node = ListNode(num) over = 0 if node.val > 9: over = 1 node.val -= 10 node.next = self.addTwoNumbers(next1, next2, over) return node
add-two-numbers
Python faster than 100% and easy to read
mpSchrader
2
595
add two numbers
2
0.398
Medium
70
https://leetcode.com/problems/add-two-numbers/discuss/2642238/Python-O(n)-Time-Space-Solution
class Solution: # Time: O(n) and Space: O(n) def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: dummy = cur = ListNode() # Creating a new node to store l1 + l2 values, dummy will give us head address &amp; cur will be used to append new nodes carry = 0 while l1 or l2 or carry: # we will continue till either one of these have positive values left v1 = l1.val if l1 else 0 v2 = l2.val if l2 else 0 # Addition: New digits val = v1 + v2 + carry # in 1st iteration carry will be 0, but in next iterations previous iter generated carry will be added carry = val // 10 # if val = 15 then 15//10 = 1 ie. (1.5) val = val % 10 # 15 % 10 = 5 cur.next = ListNode(val) # dummy/cur --> val # Update pointers cur = cur.next # dummy --> val/cur l1 = l1.next if l1 else None l2 = l2.next if l2 else None return dummy.next
add-two-numbers
Python O(n) Time-Space Solution
DanishKhanbx
1
221
add two numbers
2
0.398
Medium
71
https://leetcode.com/problems/add-two-numbers/discuss/1957325/python3-or-99.16-faster-and-O(n)-time-complexity
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: dummy = curr = ListNode(0) carry = 0 while l1 and l2 : tsum = l1.val + l2.val + carry num = tsum % 10 curr.next = ListNode(num) curr = curr.next carry = tsum // 10 l1 = l1.next l2 = l2.next while l1: tsum = l1.val + carry if tsum > 9: num = tsum % 10 curr.next = ListNode(num) carry = tsum // 10 else: curr.next = ListNode(tsum) carry = 0 curr = curr.next l1 = l1.next while l2: tsum = l2.val + carry if tsum > 9: num = tsum % 10 curr.next = ListNode(num) carry = tsum // 10 else: curr.next = ListNode(tsum) carry = 0 curr = curr.next l2 = l2.next if carry > 0: curr.next = ListNode(carry) return dummy.next
add-two-numbers
python3 | 99.16% faster and O(n) time complexity
sanketsans
1
187
add two numbers
2
0.398
Medium
72
https://leetcode.com/problems/add-two-numbers/discuss/1836620/Python-Simple-Solution
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: res = 0 ans = ListNode() cur = ans while l1 and l2: d = l1.val + l2.val + res res = d//10 cur.next = ListNode(d%10) cur = cur.next l1 = l1.next l2 = l2.next cur1 = l2 if l1: cur1 = l1 while cur1: d = cur1.val + res res = d//10 cur.next = ListNode(d%10) cur = cur.next cur1 = cur1.next if res: cur.next = ListNode(res) return ans.next
add-two-numbers
[Python] Simple Solution
zouhair11elhadi
1
65
add two numbers
2
0.398
Medium
73
https://leetcode.com/problems/add-two-numbers/discuss/1761128/Easy-to-Understand-Python-Solution
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: vals1 = [] vals2 = [] cur = dummy = ListNode(0) while l1 is not None: vals1.append(str(l1.val)) l1 = l1.next while l2 is not None: vals2.append(str(l2.val)) l2 = l2.next vals1.reverse() vals2.reverse() c = int("".join(vals1)) d = int("".join(vals2)) amount = list(str(c + d)) amount.reverse() for i in amount: cur.next = ListNode(int(i)) cur = cur.next return dummy.next
add-two-numbers
Easy to Understand Python Solution
newcomerlearn
1
304
add two numbers
2
0.398
Medium
74
https://leetcode.com/problems/add-two-numbers/discuss/1757973/Brief-and-pythonic
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: def lstIter(lst): while lst is not None: yield lst.val lst = lst.next head = prev = ListNode(0) carry = 0 for x in (ListNode(sum(x)) for x in zip_longest(lstIter(l1), lstIter(l2), fillvalue=0)): prev.next = prev = x x.val, carry = (carry + x.val) % 10, (carry + x.val) // 10 if carry: prev.next = ListNode(carry) return head.next
add-two-numbers
Brief and pythonic
Lrnx
1
110
add two numbers
2
0.398
Medium
75
https://leetcode.com/problems/add-two-numbers/discuss/1582607/Python-Solution
class Solution: def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]: output = curr = ListNode() o = 0 #overflow while l1 or l2: n1, n2 = 0, 0 if l1: n1 = l1.val l1 = l1.next if l2: n2 = l2.val l2 = l2.next if n1 + n2 + o >= 10: curr.next = ListNode(n1+n2+o-10) o = 1 else: curr.next = ListNode(n1+n2+o) o = 0 curr = curr.next if o: curr.next = ListNode(o) return output.next
add-two-numbers
Python Solution
overzh_tw
1
178
add two numbers
2
0.398
Medium
76
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/742926/Simple-Explanation-or-Concise-or-Thinking-Process-and-Example
class Solution(object): def lengthOfLongestSubstring(self, s): """ :type s: str :rtype: int abcabcbb """ if len(s) == 0: return 0 seen = {} left, right = 0, 0 longest = 1 while right < len(s): if s[right] in seen: left = max(left,seen[s[right]]+1) longest = max(longest, right - left + 1) seen[s[right]] = right right += 1 print(left, right, longest) return longest
longest-substring-without-repeating-characters
Simple Explanation | Concise | Thinking Process & Example
ivankatrump
290
13,100
longest substring without repeating characters
3
0.338
Medium
77
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2132954/Python-Simple-Solution-w-Explanation-or-Brute-Force-%2B-Sliding-Window
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: res = 0 seen = set() for start_idx in range(len(s)): seen.clear() end_idx = start_idx while end_idx < len(s): if s[end_idx] in seen: break seen.add(s[end_idx]) end_idx += 1 res = max(res, end_idx - start_idx) return res
longest-substring-without-repeating-characters
✅ [Python] Simple Solution w/ Explanation | Brute-Force + Sliding Window
r0gue_shinobi
26
1,900
longest substring without repeating characters
3
0.338
Medium
78
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2132954/Python-Simple-Solution-w-Explanation-or-Brute-Force-%2B-Sliding-Window
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: prev = [-1] * 128 res, start_idx = 0, 0 for end_idx, char in enumerate(s): if prev[ord(char)] >= start_idx: start_idx = prev[ord(char)] + 1 prev[ord(char)] = end_idx res = max(res, end_idx - start_idx + 1) return res
longest-substring-without-repeating-characters
✅ [Python] Simple Solution w/ Explanation | Brute-Force + Sliding Window
r0gue_shinobi
26
1,900
longest substring without repeating characters
3
0.338
Medium
79
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2132791/Python-Easy-2-approaches
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: counter = defaultdict(int) # track counts of each character l=0 max_length=0 for r, c in enumerate(s): counter[c]+=1 if counter[c] > 1: while l<r and counter[c]>1: # iterate until window is valid counter[s[l]]-=1 l+=1 max_length=max(max_length, r-l+1) return max_length
longest-substring-without-repeating-characters
Python Easy 2 approaches ✅
constantine786
21
2,000
longest substring without repeating characters
3
0.338
Medium
80
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2132791/Python-Easy-2-approaches
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: last_seen = {} l=0 max_length=0 for r in range(len(s)): if s[r] in last_seen: l=max(last_seen[s[r]], l) last_seen[s[r]]=r+1 max_length=max(max_length, r-l+1) return max_length
longest-substring-without-repeating-characters
Python Easy 2 approaches ✅
constantine786
21
2,000
longest substring without repeating characters
3
0.338
Medium
81
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2799840/Python-or-Easy-Solution
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: output = 0 count = {} pos = -1 for index, letter in enumerate(s): if letter in count and count[letter] > pos: pos = count[letter] count[letter] = index output = max(output,index-pos) return output
longest-substring-without-repeating-characters
Python | Easy Solution✔
manayathgeorgejames
18
2,800
longest substring without repeating characters
3
0.338
Medium
82
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1526581/Simplest-way-(with-explanation)-97-faster
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: string = s max_length = 0 # we set max_length to 0 because string may be empty. seen_character = '' # a empty string to store the character that we have already seen. for letter in string: # we are checking every letter/character in string... if letter not in seen_character: seen_character += letter # if letter not in there then we add to it. else: ## now if the letter is already in seen_character then we get the index of that letter by using seen_character.index() and then we slice the string from that index+1 to last, so that the the first seen letter will be removed. # for example - 'abcabbd' # here after 'abc' , again "a" was there so we get the index of first "a" and slice the string then be get string = "bc" . seen_character = seen_character[seen_character.index(letter) + 1:] + letter # and then we add the letter "a" to the last. so the string will become "bca" max_length = max(max_length, len(seen_character)) # here we use a function max() that everytime return the maximum value between two number. it sets max_length each time the loop runs. return max_length # finally return the maximum length. #by - Tony Stark
longest-substring-without-repeating-characters
Simplest way (with explanation) [97% faster]
tony_stark_47
16
1,000
longest substring without repeating characters
3
0.338
Medium
83
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2219528/Simple-Python-Solution-oror-Sliding-Window
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: set_ = set() res = 0 l = 0 for r in range(len(s)): while s[r] in set_: set_.remove(s[l]) l += 1 set_.add(s[r]) res = max(res, r-l+1) return res # An Upvote will be encouraging
longest-substring-without-repeating-characters
Simple Python Solution || Sliding Window
rajkumarerrakutti
13
687
longest substring without repeating characters
3
0.338
Medium
84
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1074783/Python-Interview-Thought-Process%3A-O(2n)-greater-O(n)
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: queue = collections.deque([]) window = set() result = 0 for c in s: if c in window: while queue: prev = queue.popleft() window.remove(prev) if prev == c: break queue.append(c) window.add(c) result = max(result, len(window)) return result
longest-substring-without-repeating-characters
Python Interview Thought Process: O(2^n) -> O(n)
dev-josh
11
1,300
longest substring without repeating characters
3
0.338
Medium
85
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1953837/Python-Easiest-Solution-With-Explanation-or-92.68-Fasteror-Beg-to-advor-Sliding-Window
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: left = 0 res = 0 charSet = set() # taking set to have unique values. for right in range(len(s)): while s[right] in charSet: # if we are getting duplication character then we have to update our set. charSet.remove(s[left]) left+=1 charSet.add(s[right]) # once duplicate value is removed we are going to add right most element to the set res = max(res,right-left+1) # at this stage we are sure that we dont have any duplicate , so lets update the res variable now. return res
longest-substring-without-repeating-characters
Python Easiest Solution With Explanation | 92.68 % Faster| Beg to adv| Sliding Window
rlakshay14
9
362
longest substring without repeating characters
3
0.338
Medium
86
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1431722/Python-easy-solution-using-hashMap
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: if s is None: return 0 if len(s) <= 1: return len(s) charMap = dict() start = 0 longest = 0 for i,c in enumerate(s): if c in charMap: start = max(start, charMap[c]+1) longest = max(longest, i-start+1) charMap[c] = i return longest
longest-substring-without-repeating-characters
Python easy solution using hashMap
bTest2
9
1,100
longest substring without repeating characters
3
0.338
Medium
87
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1325444/easiest-solution-python-3......fastest
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: start = 0 end = 0 max_len = 0 d={} while end<len(s): if s[end] in d and d[s[end]] >= start: start = d[s[end]]+1 max_len = max(max_len , end-start+1) d[s[end]] = end end+=1 return(max_len)
longest-substring-without-repeating-characters
easiest solution python 3......fastest
pravishbajpai06
9
2,100
longest substring without repeating characters
3
0.338
Medium
88
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2166571/Python-Sliding-Window-(Thought-Process)
class Solution: def lengthOfLongestSubstring(self, str: str) -> int: # Handle empty input if not str: return 0 # Define result, start/end pointers, hashmap for seen characters length = 1 start = 0 end = 0 seen = {} # Iterate through string using sliding window technique while end < len(str): # You don't have to do this, but slightly cleaner startChar = str[start] endChar = str[end] # If our end character has already been seen... if endChar in seen: # We should reset our start to the new end (+1), or the new start (if our last seen "end" char is before our current start) start = max(start, seen[endChar] + 1) # We set the length of our longest known substring w/out repeating characters length = max(length, end - start + 1) # We reset the index we've last seen end char at (or add it, if never seen before) seen[endChar] = end # Expand our window end += 1 # Return our longest substring w/ no repeating characters return length
longest-substring-without-repeating-characters
[Python] Sliding Window (Thought Process)
ernestshackleton
8
424
longest substring without repeating characters
3
0.338
Medium
89
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1245216/Python-32ms-simple-and-easy
class Solution: def lengthOfLongestSubstring(self, s): unique="" a=0 for c in s: if c not in unique: unique+=c else: unique = unique[unique.index(c)+1:]+c a = max(a,len(unique)) return a
longest-substring-without-repeating-characters
Python- 32ms simple and easy
akashadhikari
8
664
longest substring without repeating characters
3
0.338
Medium
90
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1595492/Python3-40-ms-faster-than-99.69
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: res = 0 longest_substr = '' for ch in s: if ch not in longest_substr: longest_substr += ch if len(longest_substr) > res: res += 1 else: i = longest_substr.find(ch) longest_substr = longest_substr[i+1:] + ch return res
longest-substring-without-repeating-characters
[Python3] 40 ms, faster than 99.69%
larysa_sh
6
576
longest substring without repeating characters
3
0.338
Medium
91
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1256739/Python3-simple-solution-using-sliding-window-and-set
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: left = right = ans = 0 char = set() res = 0 while right < len(s): if s[right] not in char: char.add(s[right]) right += 1 else: char.remove(s[left]) res = max(res, right-left) left += 1 res = max(res, right - left) return res
longest-substring-without-repeating-characters
Python3 simple solution using sliding-window and set
EklavyaJoshi
6
342
longest substring without repeating characters
3
0.338
Medium
92
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/240337/Python3-Simple-O(n)-Solution-68ms-runtime(99.52)-and-memory(100)
class Solution: def lengthOfLongestSubstring(self, s: 'str') -> 'int': max_sequence = "" current_sequence = "" for ch in s: if ch in current_sequence: if len(current_sequence) > len(max_sequence): max_sequence = current_sequence current_sequence = current_sequence[current_sequence.index(ch) + 1:] + ch else: current_sequence += ch return max(len(max_sequence), len(current_sequence))
longest-substring-without-repeating-characters
Python3 - Simple O(n) Solution 68ms, runtime(99.52%) and memory(100%)
nikhil_g777
6
559
longest substring without repeating characters
3
0.338
Medium
93
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1453922/Python3-Hashmap-easy-solution-or-faster-than-99.94
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: start = -1 # start index of current substring longest = 0 # length of the longest substring hash_map = dict() # hash map is to store the latest index of char for i in range(len(s)): if s[i] in hash_map and start < hash_map[s[i]]: start = hash_map[s[i]] hash_map[s[i]] = i if i - start > longest: longest = i - start return longest
longest-substring-without-repeating-characters
Python3 Hashmap easy solution | faster than 99.94%
Janetcxy
5
607
longest substring without repeating characters
3
0.338
Medium
94
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2283305/Python-Sliding-Window-O(n)-beats-96.91-(Explanation)
class Solution(object): def lengthOfLongestSubstring(self, s): mostRecentIndexofChar = {} longest = 0 firstGoodIndex = 0 for index in range(len(s)): if firstGoodIndex > index: continue if s[index] in mostRecentIndexofChar: firstGoodIndex = max(firstGoodIndex, mostRecentIndexofChar[s[index]]) longest = max(longest, index - firstGoodIndex + 1) mostRecentIndexofChar[s[index]] = index + 1 return longest
longest-substring-without-repeating-characters
[Python] Sliding Window O(n) beats 96.91% (Explanation)
Derek_Shimoda
4
356
longest substring without repeating characters
3
0.338
Medium
95
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/2133403/Python3-WINDOW-with-SET-(-*).-Explained
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: hset, start, ans = set(), 0, 0 for ch in s: while ch in hset: hset.remove(s[start]) start += 1 hset.add(ch) ans = max(ans, len(hset)) return max(ans, len(hset))
longest-substring-without-repeating-characters
✔️ [Python3] WINDOW with SET (^-^*)/. Explained
artod
4
153
longest substring without repeating characters
3
0.338
Medium
96
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1788064/Python3-oror-Strings-oror-98-Faster
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: max_length = 0 non_rep_sub_str = "" for i in s: if i in non_rep_sub_str: non_rep_sub_str = non_rep_sub_str.split(i)[1] + i else: non_rep_sub_str += i max_length = max(max_length, len(non_rep_sub_str)) return max_length
longest-substring-without-repeating-characters
Python3 || Strings || 98% Faster
cherrysri1997
4
396
longest substring without repeating characters
3
0.338
Medium
97
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1621748/Python3-Sliding-Window-%2B-Hashmap-or-9-Lines-Short-or-O(n)-Time-and-Space
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: maxSize = l = 0 window = {} for r, char in enumerate(s): l = window[char] + 1 if char in window and window[char] >= l else l window[char] = r maxSize = max(maxSize, r - l + 1) return maxSize
longest-substring-without-repeating-characters
[Python3] Sliding Window + Hashmap | 9 Lines Short | O(n) Time & Space
PatrickOweijane
4
701
longest substring without repeating characters
3
0.338
Medium
98
https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1539793/Python-Runtime-52-ms-94%2B%2B-Faster-Solution
class Solution: def lengthOfLongestSubstring(self, s: str) -> int: key = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z', '!', ' ','0','1','2', '3', '4', '5', '6', '7', '8', '9', 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z', '!', '"', '#', '$', '%', '&amp;', "'", '(', ')', '*', '+', ',', '-', '.', '/', ':', ';', '<', '=', '>', '?', '@', '[', '\\', ']', '^', '_', '`', '{', '|', '}', '~'] value = True counter = dict.fromkeys(key, value) i = j = r = 0 while True: try: if counter.get(s[j]): counter[s[j]] = False j += 1 else: counter[s[i]] = True l = j - i i += 1 if l > r: r = l except: l = j - i return l if l > r else r
longest-substring-without-repeating-characters
Python Runtime 52 ms 94%++ Faster Solution
aaffriya
4
697
longest substring without repeating characters
3
0.338
Medium
99
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