post_href
stringlengths
57
213
python_solutions
stringlengths
71
22.3k
slug
stringlengths
3
77
post_title
stringlengths
1
100
user
stringlengths
3
29
upvotes
int64
-20
1.2k
views
int64
0
60.9k
problem_title
stringlengths
3
77
number
int64
1
2.48k
acceptance
float64
0.14
0.91
difficulty
stringclasses
3 values
__index_level_0__
int64
0
34k
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1661711/Python3-or-O(n)-time-complexity-or-Easy-to-Understand
class Solution: def replaceDigits(self, s: str) -> str: if not len(s): return s alpha_list = ["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"] output_str = "" for i in range(1,len(s),2): #iter...
replace-all-digits-with-characters
Python3 | O(n) time complexity | Easy to Understand
Meow1303
0
78
replace all digits with characters
1,844
0.798
Easy
26,300
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1554196/EASY-AND-SIMPLE-SOLN-(faster-than-99)
class Solution: def replaceDigits(self, s: str) -> str: string,i,j = "abcdefghijklmnopqrstuvwxyz",0,1 L = list(s) while j < len(s): idx = string.index(s[i]) L[j] = string[idx+int(s[j])] i += 2 j += 2 return "".join([i for...
replace-all-digits-with-characters
EASY AND SIMPLE SOLN (faster than 99%)
anandanshul001
0
85
replace all digits with characters
1,844
0.798
Easy
26,301
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1487311/1-line-solution-in-Python
class Solution: def replaceDigits(self, s: str) -> str: return "".join(chr(ord(s[i - 1]) + int(s[i])) if i &amp; 1 else s[i] for i in range(len(s)))
replace-all-digits-with-characters
1-line solution in Python
mousun224
0
94
replace all digits with characters
1,844
0.798
Easy
26,302
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1369024/Python3-ASCII-Solution
class Solution: def replaceDigits(self, s: str) -> str: retstring = "" for i in range(0, len(s)): if not s[i].isdigit(): retstring += s[i] else: prev_ascii = ord(s[i - 1]) new_ascii = prev_ascii + int(s[i]) to_l...
replace-all-digits-with-characters
Python3 ASCII Solution
RobertObrochta
0
120
replace all digits with characters
1,844
0.798
Easy
26,303
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1277278/Simple-Python-Solution-or-O(N)-time-or-O(N)-space
class Solution: def replaceDigits(self, s: str) -> str: def shift(d, n): return chr(ord(d) + int(n)) answer = "" for i in range(len(s)): if i % 2 == 0: answer += s[i] else: answer += shift(s[i - 1], s[i]) return answer
replace-all-digits-with-characters
Simple Python Solution | O(N) time | O(N) space
vanigupta20024
0
159
replace all digits with characters
1,844
0.798
Easy
26,304
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1241798/python-or-easy-solution
class Solution: def replaceDigits(self, s: str) -> str: u='abcdefghijklmnopqrstuvwxyz' s=list(s) for j in range(0,len(s),2): if j+1<len(s): s[j+1] = u[(u.index(s[j])+int(s[j+1]))%26] return ''.join(s)
replace-all-digits-with-characters
python | easy solution
chikushen99
0
100
replace all digits with characters
1,844
0.798
Easy
26,305
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1236696/PYTHON-CONVERT-TO-ASCII-VALUE-AND-AGAIN-TO-CHARACTER
class Solution: def replaceDigits(self, s: str) -> str: res="" for i in range(len(s)): try: j = int(s[i]) res+=chr(ord(s[i-1]) + j) except: print(s[i]) res+=s[i] ...
replace-all-digits-with-characters
PYTHON -- CONVERT TO ASCII VALUE AND AGAIN TO CHARACTER
shubhamdec10
0
60
replace all digits with characters
1,844
0.798
Easy
26,306
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1191307/Python3-linear-sweep
class Solution: def replaceDigits(self, s: str) -> str: s = list(s) for i in range(1, len(s), 2): s[i] = chr(ord(s[i-1]) + int(s[i])) return "".join(s)
replace-all-digits-with-characters
[Python3] linear sweep
ye15
0
76
replace all digits with characters
1,844
0.798
Easy
26,307
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1187880/Python3-simple-solution-beat-100-users
class Solution: def replaceDigits(self, s: str) -> str: x = '' for i in range(len(s)): if i % 2 == 0: x += s[i] else: x += chr(ord(s[i-1]) + int(s[i])) return x
replace-all-digits-with-characters
Python3 simple solution beat 100% users
EklavyaJoshi
0
46
replace all digits with characters
1,844
0.798
Easy
26,308
https://leetcode.com/problems/replace-all-digits-with-characters/discuss/1186767/Python3-100-simple-understandable-solution
class Solution: def replaceDigits(self, s: str) -> str: o='' for i in range(0,len(s),2): if i!= len(s)-1: o+=s[i]+chr((ord(s[i])+int(s[i+1]))) else: o+=s[i] return o
replace-all-digits-with-characters
Python3 100% simple understandable solution
coderash1998
0
43
replace all digits with characters
1,844
0.798
Easy
26,309
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/1503918/Python-O(N)-Time-and-Space.-Easy-to-understand
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: counter = collections.Counter(arr) available = sum(n > len(arr) for n in arr) i = ans = len(arr) while i > 0: # This number is not in arr if not counter[i]: ...
maximum-element-after-decreasing-and-rearranging
[Python] O(N) Time and Space. Easy to understand
JummyEgg
1
205
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,310
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/1185914/Pyhotn3-Sorting-Solution
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: arr.sort() arr[0] = 1 for i in range(1,len(arr)): arr[i] = min(arr[i-1]+1,arr[i]) return max(arr)
maximum-element-after-decreasing-and-rearranging
Pyhotn3 Sorting Solution
swap2001
1
36
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,311
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/2749012/python-using-sorting
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: arr.sort() c = 1 for i in range(0 , len(arr)): if c <= arr[i]: c += 1 return c - 1
maximum-element-after-decreasing-and-rearranging
python using sorting
akashp2001
0
5
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,312
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/2321142/Only-Sort-or-Python-3
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: arr.sort() if arr[0]!= 1: arr[0] = 1 maxi = arr[0] print(arr) for i in range(1,len(arr)): if abs(arr[i]-arr[i-1])>1: arr[i] = arr[i...
maximum-element-after-decreasing-and-rearranging
Only Sort | Python 3
user7457RV
0
31
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,313
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/1949481/Sorting-and-one-pass-98-speed
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: arr.sort() ans = 1 for n in arr[1:]: if n > ans: ans += 1 return ans
maximum-element-after-decreasing-and-rearranging
Sorting and one pass, 98% speed
EvgenySH
0
48
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,314
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/1702367/Python-simple-sorting-solution
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: arr.sort() n = len(arr) if n == 1: return 1 arr[0] = 1 for i in range(n-1): arr[i+1] = min(arr[i+1], arr[i]+1) return arr[n-1]-(arr[0]-...
maximum-element-after-decreasing-and-rearranging
Python simple sorting solution
byuns9334
0
60
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,315
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/1346617/Python-simple-and-pythonic
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: arr.sort() arr[0] = 1 for index, value in enumerate(arr[1:], 1): if 1 < (value - arr[index-1]): arr[index] = arr[index-1] + 1 return arr[-1]
maximum-element-after-decreasing-and-rearranging
[Python] simple and pythonic
cruim
0
49
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,316
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/1191310/Python3-greedy
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: arr.sort() ans = 0 for x in arr: ans = min(ans+1, x) return ans
maximum-element-after-decreasing-and-rearranging
[Python3] greedy
ye15
0
38
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,317
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/1186935/Straightforward-Python3-solution-with-explanation
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: ##################################################### # Two conditions required # 1. the first element is 1 # 2. arr[i-1] <= arr[i] <= arr[i-1]+1 ###############################...
maximum-element-after-decreasing-and-rearranging
Straightforward Python3 solution with explanation
tkuo-tkuo
0
25
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,318
https://leetcode.com/problems/maximum-element-after-decreasing-and-rearranging/discuss/1186051/PythonPython3-Solution-with-explanation
class Solution: def maximumElementAfterDecrementingAndRearranging(self, arr: List[int]) -> int: arr.sort() # sort the elements if arr[0] == 1: # if arr[0] is 1 then simply give pass as it satisfies the 1st condition pass else: # else make arr[0] to 1 to satisfy condition 1 ...
maximum-element-after-decreasing-and-rearranging
Python/Python3 Solution with explanation
prasanthksp1009
0
49
maximum element after decreasing and rearranging
1,846
0.591
Medium
26,319
https://leetcode.com/problems/closest-room/discuss/1186155/Python-3-Aggregate-sorted-list-detailed-explanation-(2080-ms)
class Solution: def closestRoom(self, rooms: List[List[int]], queries: List[List[int]]) -> List[int]: ans = [0] * len(queries) # sort queries to handle largest size queries first q = deque(sorted([(size, room, i) for i, (room, size) in enumerate(queries)], key=lambda a: (-a[0], a[1]...
closest-room
[Python 3] Aggregate sorted list, detailed explanation (2080 ms)
chestnut890123
1
111
closest room
1,847
0.353
Hard
26,320
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1186862/Python3-linear-sweep
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: ans = inf for i, x in enumerate(nums): if x == target: ans = min(ans, abs(i - start)) return ans
minimum-distance-to-the-target-element
[Python3] linear sweep
ye15
9
302
minimum distance to the target element
1,848
0.585
Easy
26,321
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1204659/Python3-simple-solution-beats-90-users
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: if nums[start] == target:return 0 i, j = start-1, start+1 while j < len(nums) or i > -1: if i > -1: if nums[i] == target: return start-i ...
minimum-distance-to-the-target-element
Python3 simple solution beats 90% users
EklavyaJoshi
2
135
minimum distance to the target element
1,848
0.585
Easy
26,322
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1584649/Python-sol-faster-than-99
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: if nums[start] == target: return 0 i = start j = start left = 0 right = 0 while i < len(nums) or j > 0 : if nums[i] == target : ...
minimum-distance-to-the-target-element
Python sol faster than 99%
elayan
1
109
minimum distance to the target element
1,848
0.585
Easy
26,323
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1187500/Python-3-simple-solution
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: n = len(nums) f = start b = start idx = None while f < n or b >= 0: if f < n: if nums[f] == target: idx = f break ...
minimum-distance-to-the-target-element
Python 3 simple solution
pankaj17n
1
59
minimum distance to the target element
1,848
0.585
Easy
26,324
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/2781774/python-solution-Beats-97.33-And-memory-Beats-53.45
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: x = len(nums)+1; for i in range(0 , len(nums)): if(nums[i] == target): x = min(x ,abs(start-i)); return x;
minimum-distance-to-the-target-element
python solution Beats 97.33% And memory Beats 53.45%
seifsoliman
0
3
minimum distance to the target element
1,848
0.585
Easy
26,325
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/2699068/Python-or-Two-loops-with-early-exit
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: n = len(nums) ans = float('inf') i = start while i < n and nums[i] != target: i += 1 if i < n: ans = i - start i = start - 1 while i >= 0 and n...
minimum-distance-to-the-target-element
Python | Two loops with early exit
on_danse_encore_on_rit_encore
0
5
minimum distance to the target element
1,848
0.585
Easy
26,326
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/2660184/Optimized-Python-Approach-Without-Min-and-Abs-or-O(n)-time-O(1)-space
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: i, j = start, start + 1 while i >= 0 and j < len(nums): if nums[i] == target: return start - i if nums[j] == target: return j - start i -= 1 ...
minimum-distance-to-the-target-element
Optimized Python Approach Without Min and Abs | O(n) time, O(1) space
kcstar
0
5
minimum distance to the target element
1,848
0.585
Easy
26,327
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/2633820/Simple-one-liner-or-Python
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: return min(abs(key - start) for key, val in enumerate(nums) if val == target)
minimum-distance-to-the-target-element
Simple one liner | Python
Abhi_-_-
0
5
minimum distance to the target element
1,848
0.585
Easy
26,328
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/2054301/Python-Solution-or-O(n)-or-One-Liner-or-Min-of-List
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: return min([abs(idx - start) for idx in range(len(nums)) if nums[idx] == target])
minimum-distance-to-the-target-element
Python Solution | O(n) | One-Liner | Min of List
Gautam_ProMax
0
71
minimum distance to the target element
1,848
0.585
Easy
26,329
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/2022150/Python-Middle-Out-or-Two-Pointers-or-Start-at-the-Start-or-Clean-and-Simple
class Solution: def getMinDistance(self, nums, target, start): n = len(nums) l = r = start while l >= 0 or r < n: if l >= 0: if nums[l]==target: return start-l l -= 1 if r < n: if nums[r]==target: return r-start ...
minimum-distance-to-the-target-element
Python - Middle-Out | Two-Pointers | Start at the Start | Clean and Simple
domthedeveloper
0
50
minimum distance to the target element
1,848
0.585
Easy
26,330
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1963869/Brute-force-list-comp
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: # find the smallest absolute difference between all indicies that contain the target &amp; start indicies = [i for i, n in enumerate(nums) if n == target] dist = [abs(start - i) for i in indicies] ret...
minimum-distance-to-the-target-element
Brute force list comp
andrewnerdimo
0
28
minimum distance to the target element
1,848
0.585
Easy
26,331
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1829262/4-Lines-Python-Solution-oror-50-Faster-oror-Memory-less-than-75
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: ans=10**4 for i in range(len(nums)): if nums[i]==target and abs(i-start)<ans: ans=abs(i-start) return ans
minimum-distance-to-the-target-element
4-Lines Python Solution || 50% Faster || Memory less than 75%
Taha-C
0
43
minimum distance to the target element
1,848
0.585
Easy
26,332
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1765258/Simple-Python3-one-liner-(beats-~77)
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: return min(abs(i - start) for i in range(len(nums)) if nums[i] == target)
minimum-distance-to-the-target-element
Simple Python3 one-liner (beats ~77%)
y-arjun-y
0
40
minimum distance to the target element
1,848
0.585
Easy
26,333
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1736947/Python-Simple-or-O(n)-or-Beats-99.56-in-Memory
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: min_sum = 10**5 for i in range(len(nums)): if nums[i]==target: if abs(i-start) < min_sum: min_sum = abs(i-start) return min_sum
minimum-distance-to-the-target-element
Python Simple | O(n) | Beats 99.56% in Memory
veerbhansari
0
44
minimum distance to the target element
1,848
0.585
Easy
26,334
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1665541/Python-1-liner%3A-O(1)-Space-O(n)-Time-Almost-Never-Visits-All-Elements
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: return next(j for j in range(len(nums)) if (start+j < len(nums) and nums[start + j] == target) or (start >= j and nums[start - j] == target))
minimum-distance-to-the-target-element
Python 1-liner: O(1) Space O(n) Time, Almost Never Visits All Elements
SamRagusa
0
62
minimum distance to the target element
1,848
0.585
Easy
26,335
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1519218/Walk-from-start-to-both-directions-87-speed
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: len_nums = len(nums) for i in range(len_nums): if (start + i < len_nums and nums[start + i] == target or start - i > - 1 and nums[start - i] == target): return i...
minimum-distance-to-the-target-element
Walk from start to both directions, 87% speed
EvgenySH
0
56
minimum distance to the target element
1,848
0.585
Easy
26,336
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1389412/Python-simple-solution
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: l = len(nums) an=[] for i in range(0,l): if nums[i] == target: an.append(abs(i-start)) return min(an)
minimum-distance-to-the-target-element
Python simple solution
kdevharsh2001
0
58
minimum distance to the target element
1,848
0.585
Easy
26,337
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1192674/Intuitive-approach-by-list-comprehension-map-and-sort
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: # 0): nums = [1,2,3,4,5], target=5, start=3 # 1): plist = [4] # 2): dlist = [1] # 3): ans = dlist[0] = 1 plist = [i for i, v in enumerate(nums) if v == target] dlist = list(map(...
minimum-distance-to-the-target-element
Intuitive approach by list comprehension, map and sort
puremonkey2001
0
37
minimum distance to the target element
1,848
0.585
Easy
26,338
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1191568/Simple-python-99-faster
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: min_abs = float('inf') for i in range(len(nums)): if nums[i]==target: min_abs = min(abs(i-start),min_abs) return min_abs
minimum-distance-to-the-target-element
Simple python 99% faster
pheobhe
0
69
minimum distance to the target element
1,848
0.585
Easy
26,339
https://leetcode.com/problems/minimum-distance-to-the-target-element/discuss/1187113/PythonPython3-Solution
class Solution: def getMinDistance(self, nums: List[int], target: int, start: int) -> int: mini = 10**9 for i,num in enumerate(nums): if num == target: mini = min(mini,abs(i-start)) return mini
minimum-distance-to-the-target-element
Python/Python3 Solution
prasanthksp1009
0
70
minimum distance to the target element
1,848
0.585
Easy
26,340
https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/discuss/2632013/Clear-Python-DFS-with-comments
class Solution: def splitString(self, s: str) -> bool: """ Time = O(2^N) Space = O(N) space from stack """ def dfs(index: int, last: int) -> bool: if index == len(s): return True # j: [index, len(s)-1] ...
splitting-a-string-into-descending-consecutive-values
Clear Python DFS with comments
changyou1009
0
8
splitting a string into descending consecutive values
1,849
0.323
Medium
26,341
https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/discuss/1837563/Slow-but-easy-solution
class Solution: def splitString(self, s: str) -> bool: s = str(int(s)) def check(l): if len(l)<2: return False #print(l) for i in range(len(l)-1): if int(l[i])!=int(l[i+1])+1: return False return Tru...
splitting-a-string-into-descending-consecutive-values
Slow but easy solution
aazad20
0
56
splitting a string into descending consecutive values
1,849
0.323
Medium
26,342
https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/discuss/1755227/Python3-or-Recursive-solution-with-a-helper-function-or-24ms-beats-99.24
class Solution: def splitString(self, s: str) -> bool: self.res = False def helper(s, k): n1 = int(s[:k+1]) for j in range(k+1, len(s)): n2 = int(s[k+1:j+1]) if n1 - n2 == 1: if j == len(s)-1: self.re...
splitting-a-string-into-descending-consecutive-values
Python3 | Recursive solution with a helper function | 24ms beats 99.24%
nandhakiran366
0
39
splitting a string into descending consecutive values
1,849
0.323
Medium
26,343
https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/discuss/1357639/String-manipulation-96-speed
class Solution: def splitString(self, s: str) -> bool: s = s.lstrip("0") if s: for end in range(1, len(s)): next_num = str(int(s[:end]) - 1) remaining_s = s[end:] if set(remaining_s) == {"0"}: remaining_s = "0" ...
splitting-a-string-into-descending-consecutive-values
String manipulation, 96% speed
EvgenySH
0
66
splitting a string into descending consecutive values
1,849
0.323
Medium
26,344
https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/discuss/1188042/Python3-Recursion-solution-for-reference
class Solution: def splitString(self, s: str) -> bool: # convert the string into digits. s = [int(i) for i in list(s)] def r(base, index, nums): ## increment power of 10. d = 0 ## accumulate the value for 10. acc = 0 res = False ## chec...
splitting-a-string-into-descending-consecutive-values
[Python3] Recursion solution for reference
vadhri_venkat
0
51
splitting a string into descending consecutive values
1,849
0.323
Medium
26,345
https://leetcode.com/problems/splitting-a-string-into-descending-consecutive-values/discuss/1186877/Python3-backtracking
class Solution: def splitString(self, s: str) -> bool: def fn(i, x): """Return True if s[i:] can be split following x.""" if i == len(s): return True if x == 0: return False ans = False for ii in range(i, len(s) - int(i == 0)): ...
splitting-a-string-into-descending-consecutive-values
[Python3] backtracking
ye15
0
24
splitting a string into descending consecutive values
1,849
0.323
Medium
26,346
https://leetcode.com/problems/minimum-adjacent-swaps-to-reach-the-kth-smallest-number/discuss/1186887/Python3-brute-force
class Solution: def getMinSwaps(self, num: str, k: int) -> int: num = list(num) orig = num.copy() for _ in range(k): for i in reversed(range(len(num)-1)): if num[i] < num[i+1]: ii = i+1 while ii < len(num) and n...
minimum-adjacent-swaps-to-reach-the-kth-smallest-number
[Python3] brute-force
ye15
7
803
minimum adjacent swaps to reach the kth smallest number
1,850
0.719
Medium
26,347
https://leetcode.com/problems/minimum-adjacent-swaps-to-reach-the-kth-smallest-number/discuss/1467322/Python-3-or-Permutation-Brute-Force-or-Explanation
class Solution: def getMinSwaps(self, num: str, k: int) -> int: def next_permutation(nums): small = len(nums) - 2 while small >= 0 and nums[small] >= nums[small+1]: small -= 1 # find last place there is an increase if small == -1: nums.reverse() ...
minimum-adjacent-swaps-to-reach-the-kth-smallest-number
Python 3 | Permutation, Brute Force | Explanation
idontknoooo
2
378
minimum adjacent swaps to reach the kth smallest number
1,850
0.719
Medium
26,348
https://leetcode.com/problems/minimum-adjacent-swaps-to-reach-the-kth-smallest-number/discuss/1928733/Python-or-using-next_perm-with-explanation
class Solution: def getMinSwaps(self, num: str, k: int) -> int: #find kth target=num for i in range(k): target=self.next_perm(target) #count step res=0 num=list(num) for i in range(len(num)): if num[i]!=target[i]: j=i #...
minimum-adjacent-swaps-to-reach-the-kth-smallest-number
Python | using next_perm with explanation
ginaaunchat
1
218
minimum adjacent swaps to reach the kth smallest number
1,850
0.719
Medium
26,349
https://leetcode.com/problems/minimum-adjacent-swaps-to-reach-the-kth-smallest-number/discuss/1431107/O(n*(k%2Bn-log-n))-T-or-O(n)-S-or-implicit-treap-%2B-next-permutation-or-Python-3
class Solution: def getMinSwaps(self, num: str, k: int) -> int: import random class ImplicitTreap: class ImplicitTreapNode: def __init__(self, y, value, left=None, right=None): self.y = y self.value = value ...
minimum-adjacent-swaps-to-reach-the-kth-smallest-number
O(n*(k+n log n)) T | O(n) S | implicit treap + next permutation | Python 3
CiFFiRO
0
167
minimum adjacent swaps to reach the kth smallest number
1,850
0.719
Medium
26,350
https://leetcode.com/problems/minimum-adjacent-swaps-to-reach-the-kth-smallest-number/discuss/1188750/Python3-Solution-for-reference-combining-next-permutation-and-adjacent-swapping
class Solution: def getMinSwaps(self, num: str, k: int) -> int: nums = [int(i) for i in num] orig = [int(i) for i in num] right = len(nums) - 1 ans = 0 for _ in range(k): for x in range(right, 0, -1): if nums[x] > nums[x-1]: it...
minimum-adjacent-swaps-to-reach-the-kth-smallest-number
[Python3] Solution for reference combining next permutation and adjacent swapping
vadhri_venkat
0
263
minimum adjacent swaps to reach the kth smallest number
1,850
0.719
Medium
26,351
https://leetcode.com/problems/minimum-interval-to-include-each-query/discuss/1422509/For-Beginners-oror-Easy-Approach-oror-Well-Explained-oror-Clean-and-Concise
class Solution: def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]: intervals.sort(key = lambda x:x[1]-x[0]) q = sorted([qu,i] for i,qu in enumerate(queries)) res=[-1]*len(queries) for left,right in intervals: ind = bisect.bisect(q,[left]) while ind...
minimum-interval-to-include-each-query
📌📌 For Beginners || Easy-Approach || Well-Explained || Clean & Concise 🐍
abhi9Rai
4
274
minimum interval to include each query
1,851
0.479
Hard
26,352
https://leetcode.com/problems/minimum-interval-to-include-each-query/discuss/2652142/Clean-Python3-or-Sorting-and-Heap
class Solution: def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]: queries_asc = sorted((q, i) for i, q in enumerate(queries)) intervals.sort() i, num_intervals = 0, len(intervals) size_heap = [] # (size, left) for pos, qnu...
minimum-interval-to-include-each-query
Clean Python3 | Sorting & Heap
ryangrayson
2
56
minimum interval to include each query
1,851
0.479
Hard
26,353
https://leetcode.com/problems/minimum-interval-to-include-each-query/discuss/1187214/PythonPython3-solution-using-heap-and-sorting-method
class Solution: def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]: # To store the output result lis = [0 for i in range(len(queries))] #sort the intervals in the reverse order intervals.sort(reverse = True) #View the intervals list print(intervals) #s...
minimum-interval-to-include-each-query
Python/Python3 solution using heap and sorting method
prasanthksp1009
2
112
minimum interval to include each query
1,851
0.479
Hard
26,354
https://leetcode.com/problems/minimum-interval-to-include-each-query/discuss/1186916/Python3-priority-queue
class Solution: def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]: intervals.sort() pq = [] k = 0 ans = [-1] * len(queries) for query, i in sorted(zip(queries, range(len(queries)))): while k < len(intervals) and intervals[k][0] <...
minimum-interval-to-include-each-query
[Python3] priority queue
ye15
2
81
minimum interval to include each query
1,851
0.479
Hard
26,355
https://leetcode.com/problems/minimum-interval-to-include-each-query/discuss/2315278/Python-3-oror-hashMap-priority-queue-min-heap
class Solution: def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]: hashMap = {} intervals.sort() minHeap = [] i_l = len(intervals) i = 0 for q in sorted(queries): while i < i_l and intervals[i][0] <= q: start, end = intervals[i] heapq.heappush(minHeap, [(end-st...
minimum-interval-to-include-each-query
Python 3 || hashMap, priority queue, min-heap
sagarhasan273
1
114
minimum interval to include each query
1,851
0.479
Hard
26,356
https://leetcode.com/problems/minimum-interval-to-include-each-query/discuss/1187869/Python-3-Dynamic-heap-with-detailed-explanation-(1900ms)
class Solution: def minInterval(self, intervals: List[List[int]], queries: List[int]) -> List[int]: # sort queries from small to large q = deque(sorted([(x, i) for i, x in enumerate(queries)])) # answer to queries, initial state set to -1 ans = [-1] * len(queries) ...
minimum-interval-to-include-each-query
[Python 3] Dynamic heap with detailed explanation (1900ms)
chestnut890123
0
78
minimum interval to include each query
1,851
0.479
Hard
26,357
https://leetcode.com/problems/maximum-population-year/discuss/1210686/Python3-O(N)
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: # the timespan 1950-2050 covers 101 years delta = [0] * 101 # to make explicit the conversion from the year (1950 + i) to the ith index conversionDiff = 1950 for l in logs: # the log's first entry, birth, ...
maximum-population-year
Python3 O(N)
signifying
21
2,000
maximum population year
1,854
0.599
Easy
26,358
https://leetcode.com/problems/maximum-population-year/discuss/1198702/Python3-greedy
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: vals = [] for x, y in logs: vals.append((x, 1)) vals.append((y, -1)) ans = prefix = most = 0 for x, k in sorted(vals): prefix += k if prefix > most: ...
maximum-population-year
[Python3] greedy
ye15
7
886
maximum population year
1,854
0.599
Easy
26,359
https://leetcode.com/problems/maximum-population-year/discuss/1206057/python-solution
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: b=[logs[i][0] for i in range(len(logs))] d=[logs[i][1] for i in range(len(logs))] m=0 #max population a=0 #alive r=1950 for i in range(1950,2051): a+=b.count(i)-d.count(i) ...
maximum-population-year
python solution
_Light_
2
226
maximum population year
1,854
0.599
Easy
26,360
https://leetcode.com/problems/maximum-population-year/discuss/2714948/Simple-python-solution
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: dic={} logs.sort() initial=logs[0][0] final=logs[-1][1] for i in range(initial,final): dic[i]=0 for j in logs: if(i>=j[0] and i<j[1]): dic[i]...
maximum-population-year
Simple python solution
Kiran_Rokkam
0
11
maximum population year
1,854
0.599
Easy
26,361
https://leetcode.com/problems/maximum-population-year/discuss/2560790/using-heapq
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: # sort based on birth logs.sort() # an array to save info death_birth_years = [] # maximum length maxlen = 0 # output year year = None # itearte over logs for i...
maximum-population-year
using heapq
krishnamsgn
0
24
maximum population year
1,854
0.599
Easy
26,362
https://leetcode.com/problems/maximum-population-year/discuss/2481447/Python3-differential-array
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: # Year_Converted = Year - 1949 # The converted year is between 1 and 101 diff_array = [0] * 102 for log in logs: birth_converted = log[0] - 1949 die_converted = log[1] - 1949 ...
maximum-population-year
Python3 differential array
Aden-Q
0
41
maximum population year
1,854
0.599
Easy
26,363
https://leetcode.com/problems/maximum-population-year/discuss/2021335/Python-Array-and-Dictionary-or-Clean-and-Simple!
class Solution: def maximumPopulation(self, logs): start = 1950 population = [0] * 101 for b,d in logs: b,d = b-start, d-start for year in range(b, d): population[year] += 1 return population.index(max(population)) + start
maximum-population-year
Python - Array and Dictionary | Clean and Simple!
domthedeveloper
0
94
maximum population year
1,854
0.599
Easy
26,364
https://leetcode.com/problems/maximum-population-year/discuss/2021335/Python-Array-and-Dictionary-or-Clean-and-Simple!
class Solution: def maximumPopulation(self, logs): population = Counter() for b,d in logs: for year in range(b, d): population[year] += 1 maxPopulation = max(population.values()) return min(k for k,v in population.items() if v == maxPopulation)
maximum-population-year
Python - Array and Dictionary | Clean and Simple!
domthedeveloper
0
94
maximum population year
1,854
0.599
Easy
26,365
https://leetcode.com/problems/maximum-population-year/discuss/1857572/Simple-Python-Solution-oror-55-Faster-oror-Memory-less-than-60
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: years=sorted(list(chain(*logs))) ; ans=0 ; mx=0 for year in years: cnt=0 for i in range(len(logs)): if logs[i][0]<=year and logs[i][1]>year: cnt+=1 if cnt>mx: mx=cnt ; ans=y...
maximum-population-year
Simple Python Solution || 55% Faster || Memory less than 60%
Taha-C
0
88
maximum population year
1,854
0.599
Easy
26,366
https://leetcode.com/problems/maximum-population-year/discuss/1857572/Simple-Python-Solution-oror-55-Faster-oror-Memory-less-than-60
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: d = defaultdict(int) for birth, death in logs: for i in range(birth, death): d[i] += 1 return max(d.items(), key=lambda x:(x[1],-x[0]))[0]
maximum-population-year
Simple Python Solution || 55% Faster || Memory less than 60%
Taha-C
0
88
maximum population year
1,854
0.599
Easy
26,367
https://leetcode.com/problems/maximum-population-year/discuss/1529568/One-pass-over-logs-and-years
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: years = defaultdict(int) max_population = max_year = 0 for start, end in logs: for y in range(start, end): years[y] += 1 if (years[y] > max_population or ...
maximum-population-year
One pass over logs and years
EvgenySH
0
182
maximum population year
1,854
0.599
Easy
26,368
https://leetcode.com/problems/maximum-population-year/discuss/1388920/Explained-Python3-Simulation-with-two-events-in-timeline-'born'-'dead'
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: events = [] for born, died in logs: events.extend([(born, '1_born'), (died, '0_died')]) # let us simulate the events in timeline order # the order in which they happened events...
maximum-population-year
Explained Python3 Simulation, with two events in timeline, 'born' 'dead'
yozaam
0
144
maximum population year
1,854
0.599
Easy
26,369
https://leetcode.com/problems/maximum-population-year/discuss/1229181/Python3-simple-solution-using-dictionary
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: d = {} for i,j in enumerate(logs): for k in range(j[0],j[1]): d[k] = d.get(k,0)+1 return sorted([i for i in d.keys() if d[i] == max(d.values())])[0]
maximum-population-year
Python3 simple solution using dictionary
EklavyaJoshi
-1
121
maximum population year
1,854
0.599
Easy
26,370
https://leetcode.com/problems/maximum-population-year/discuss/1223429/easy-solution-or-array-difference-algorithmor-python
class Solution: def maximumPopulation(self, logs: List[List[int]]) -> int: u=[0]*(3000) for j in logs: u[j[0]]+=1 u[j[1]]-=1 p=[u[0]] for j in range(1,len(u)): p.append(p[-1]+u[...
maximum-population-year
easy solution | array difference algorithm| python
chikushen99
-1
170
maximum population year
1,854
0.599
Easy
26,371
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/2209743/Python3-simple-solution-using-two-pointers
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: length1, length2 = len(nums1), len(nums2) i,j = 0,0 result = 0 while i < length1 and j < length2: if nums1[i] > nums2[j]: i+=1 else: resu...
maximum-distance-between-a-pair-of-values
📌 Python3 simple solution using two pointers
Dark_wolf_jss
6
69
maximum distance between a pair of values
1,855
0.527
Medium
26,372
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/1198697/Python-Solution-Simple-2-Pointer
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: max_diff = 0 i = 0 j = 0 while i<len(nums1) and j <len(nums2): if i <= j: if nums1[i] <= nums2[j]: max_diff=max(max_diff,j-i) j += 1 ...
maximum-distance-between-a-pair-of-values
Python Solution- Simple -2 Pointer
SaSha59
3
132
maximum distance between a pair of values
1,855
0.527
Medium
26,373
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/1198720/Python3-sliding-window
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: ans = ii = 0 for i, x in enumerate(nums2): while ii <= i and ii < len(nums1) and nums1[ii] > nums2[i]: ii += 1 if ii < len(nums1) and nums1[ii] <= nums2[i]: ans = max(ans, i - ii) retu...
maximum-distance-between-a-pair-of-values
[Python3] sliding window
ye15
2
68
maximum distance between a pair of values
1,855
0.527
Medium
26,374
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/1944437/Python-Binary-Search
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: result = 0 max_dist = 0 m, n = len(nums1), len(nums2) #iterate from left of nums1 for i in range(m): #for every element, find the index of nums2 with value <= nums1[i] in the range nums2[i:len(nums...
maximum-distance-between-a-pair-of-values
Python Binary Search
eforean
1
46
maximum distance between a pair of values
1,855
0.527
Medium
26,375
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/2775161/Easy-Linear-solution-O(1)-space-complexity
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: res = 0 m, n = len(nums1), len(nums2) i, j = 0, 0 while i < m and j < n: if nums1[i] <= nums2[j]: res = max(res, j-i) j += 1 else: ...
maximum-distance-between-a-pair-of-values
Easy Linear solution, O(1) space complexity
byuns9334
0
5
maximum distance between a pair of values
1,855
0.527
Medium
26,376
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/2599020/Python3-or-Solved-By-Performing-Binary-Search-on-nums2-to-find-RightMost-Limit-for-Every-index-I
class Solution: #let n = len(nums1) and m = len(nums2)! #Time-Complexity: O(n * m) #Space-Complexity: O(1) def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: #Approach: Linearly try each and every element in nums1 array! Perform binary #search to find rightmost element in ...
maximum-distance-between-a-pair-of-values
Python3 | Solved By Performing Binary Search on nums2 to find RightMost Limit for Every index I
JOON1234
0
10
maximum distance between a pair of values
1,855
0.527
Medium
26,377
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/2494547/Python-O(N)-Solution
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: i = 0 j = 0 max_distance = float("-inf") while i < len(nums1) and j < len(nums2): if nums1[i] <= nums2[j]: max_distance = max(max_distance, j-i) else: ...
maximum-distance-between-a-pair-of-values
Python O(N) Solution
yash921
0
32
maximum distance between a pair of values
1,855
0.527
Medium
26,378
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/2459975/C%2B%2BPython-Faster-Optimized-solution
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: i=0 j=0 res = 0 n = len(nums1) m = len(nums2) while i<n and j<m: if nums1[i]>nums2[j]: i+=1 else: res=max(res,j-i) j+=1 return res
maximum-distance-between-a-pair-of-values
C++/Python Faster Optimized solution
arpit3043
0
48
maximum distance between a pair of values
1,855
0.527
Medium
26,379
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/2393307/Python-Accurate-Faster-Solution-oror-Documented
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: maxD = 0 # initialize max distance with 0 i = j = 0 # initialize array pointers with 0 # traverse until both pointers < array-length while i < len(n...
maximum-distance-between-a-pair-of-values
[Python] Accurate Faster Solution || Documented
Buntynara
0
10
maximum distance between a pair of values
1,855
0.527
Medium
26,380
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/2200574/Easy-2-pointer-approach
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: i, j, n, m, maxDistance = 0, 0, len(nums1), len(nums2), 0 while i < n and j < m: if nums1[i] <= nums2[j]: maxDistance = max(maxDistance, j - i) else: i +...
maximum-distance-between-a-pair-of-values
Easy 2 pointer approach
Vaibhav7860
0
32
maximum distance between a pair of values
1,855
0.527
Medium
26,381
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/1215959/Python3-Simple-Two-Pointers-Approach-Explained-with-Comments
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: ''' 1. Start from the rightmost index of nums2 and nums1. If len(nums1)>len(nums2), start from index len(nums2)-1 of array nums1 2. Let j be the current index of nums2 and i be the current index of nums1 ...
maximum-distance-between-a-pair-of-values
Python3 Simple Two Pointers Approach, Explained with Comments
bPapan
0
72
maximum distance between a pair of values
1,855
0.527
Medium
26,382
https://leetcode.com/problems/maximum-distance-between-a-pair-of-values/discuss/1199013/Straightforward-Python3-O(n)-two-pointers-solution-with-explanation
class Solution: def maxDistance(self, nums1: List[int], nums2: List[int]) -> int: ################################################################### # Assumption/Requirement ################################################################### # nums1 and nums2 are both decreasing ...
maximum-distance-between-a-pair-of-values
Straightforward Python3 O(n) two-pointers solution with explanation
tkuo-tkuo
0
38
maximum distance between a pair of values
1,855
0.527
Medium
26,383
https://leetcode.com/problems/maximum-subarray-min-product/discuss/1198800/Python3-mono-stack
class Solution: def maxSumMinProduct(self, nums: List[int]) -> int: prefix = [0] for x in nums: prefix.append(prefix[-1] + x) ans = 0 stack = [] for i, x in enumerate(nums + [-inf]): # append "-inf" to force flush all elements while stack and stack[-1][1...
maximum-subarray-min-product
[Python3] mono-stack
ye15
10
345
maximum subarray min product
1,856
0.378
Medium
26,384
https://leetcode.com/problems/maximum-subarray-min-product/discuss/1200240/Python-monotonic-stack-beats-100
class Solution: def maxSumMinProduct(self, nums: List[int]) -> int: mod=int(1e9+7) stack=[] # (index, prefix sum at index) rsum=0 res=0 nums.append(0) for i, v in enumerate(nums): while stack and nums[stack[-1][0]] >= v: i...
maximum-subarray-min-product
Python monotonic stack beats 100%
alan9259
5
509
maximum subarray min product
1,856
0.378
Medium
26,385
https://leetcode.com/problems/maximum-subarray-min-product/discuss/2411914/Python3-or-Similar-to-Max-Rectangle-in-Histogram
class Solution: def maxSumMinProduct(self, nums: List[int]) -> int: n,ans=len(nums),-float('inf') mod=10**9+7 left=[0 for i in range(n)] right=[0 for i in range(n)] prefixSum=[0 for i in range(n)] prefixSum[0]=nums[0] for i in range(1,n): prefixSum...
maximum-subarray-min-product
[Python3] | Similar to Max Rectangle in Histogram
swapnilsingh421
0
54
maximum subarray min product
1,856
0.378
Medium
26,386
https://leetcode.com/problems/maximum-subarray-min-product/discuss/2057686/python3-dp-and-stack-solutions-for-reference.
class Solution: def dp(self, nums): N = len(nums) dp = [(0, 0) for _ in range(N)] ans = 0 for x in range(N): for y in range(x, N): if x == y: dp[y] = (nums[y], nums[y]) else: dp[y] = (min(dp[...
maximum-subarray-min-product
[python3] dp and stack solutions for reference.
vadhri_venkat
0
76
maximum subarray min product
1,856
0.378
Medium
26,387
https://leetcode.com/problems/maximum-subarray-min-product/discuss/1436397/Python-Clean
class Solution: def maxSumMinProduct(self, nums: List[int]) -> int: N, maximum = len(nums), 0 cumulative, stack = [0], [] for num in nums: cumulative.append(cumulative[-1] + num) for i, num in enumerate(nums + [-float(inf)]): start = i ...
maximum-subarray-min-product
[Python] Clean
soma28
0
223
maximum subarray min product
1,856
0.378
Medium
26,388
https://leetcode.com/problems/maximum-subarray-min-product/discuss/1203747/very-easy-python-soln
class Solution(object): def maxSumMinProduct(self, nums): n,ans,mod=len(nums),0,10**9 + 7 lft,st=[0],[[nums[0],0]] for i in range(1,n): cur=0 while st and nums[i]<=st[-1][0]: cur+=st.pop()[1]+1 st.append([nums[i],cur]) lft.appen...
maximum-subarray-min-product
very easy python soln
aayush_chhabra
0
257
maximum subarray min product
1,856
0.378
Medium
26,389
https://leetcode.com/problems/maximum-subarray-min-product/discuss/1201859/python-simple-intuitive-greedy-solution
class Solution: def maxSumMinProduct(self, nums: List[int]) -> int: n = len(nums) s = list(accumulate(nums)) # sum t = sorted([(v, i) for i, v in enumerate(nums)], reverse=True) # to greedily iterate over values j = [i for i in range(n)] # jump array, at first one can only jump in it...
maximum-subarray-min-product
[python] simple intuitive greedy solution
iamlegend123
0
154
maximum subarray min product
1,856
0.378
Medium
26,390
https://leetcode.com/problems/largest-color-value-in-a-directed-graph/discuss/1200668/easy-python-sol-..
class Solution(object): def largestPathValue(self, colors, edges): n=len(colors) graph=defaultdict(list) indegree=defaultdict(int) for u,v in edges: graph[u].append(v) indegree[v]+=1 queue=[] dp=[[0]*26 for _ in range(n)] ...
largest-color-value-in-a-directed-graph
easy python sol ..
aayush_chhabra
9
299
largest color value in a directed graph
1,857
0.407
Hard
26,391
https://leetcode.com/problems/largest-color-value-in-a-directed-graph/discuss/1199847/Python3-dp
class Solution: def largestPathValue(self, colors: str, edges: List[List[int]]) -> int: graph = {} indeg = [0] * len(colors) for u, v in edges: indeg[v] += 1 graph.setdefault(u, []).append(v) # Kahn's algo roots = [x for x in range(len(co...
largest-color-value-in-a-directed-graph
[Python3] dp
ye15
2
98
largest color value in a directed graph
1,857
0.407
Hard
26,392
https://leetcode.com/problems/largest-color-value-in-a-directed-graph/discuss/1206392/Python-Iterative-DFS-Solution
class Solution: def largestPathValue(self, colors: str, edges: List[List[int]]) -> int: d = defaultdict(list) for v1, v2 in edges: d[v1].append(v2) visited = {} counts = [[0] * 26 for _ in range(len(colors))] stack = [] global_max = 0 for root in d...
largest-color-value-in-a-directed-graph
Python Iterative DFS Solution
kylergao
0
130
largest color value in a directed graph
1,857
0.407
Hard
26,393
https://leetcode.com/problems/sorting-the-sentence/discuss/1219040/Python3-99.60-Fast-Solution
class Solution: def sortSentence(self, s: str) -> str: arr = [i[-1] + i[:-1] for i in s.split()] arr.sort() ans = "" for i in arr: ans += i[1:] + ' ' return ans[:-1]
sorting-the-sentence
[Python3] 99.60% Fast Solution
VoidCupboard
23
2,000
sorting the sentence
1,859
0.844
Easy
26,394
https://leetcode.com/problems/sorting-the-sentence/discuss/1210285/2-Lines-or-Python-Solution-or-Linear-TIme-or-Linear-Memory
class Solution: def sortSentence(self, s: str) -> str: # split the string and sort the words based upon the last letter word_list = sorted(s.split(), key = lambda word: word[-1], reverse = False) return " ".join([word[:-1] for word in word_list]) # join the words, after removing the last letter i...
sorting-the-sentence
2 Lines | Python Solution | Linear TIme | Linear Memory
ramanaditya
19
1,800
sorting the sentence
1,859
0.844
Easy
26,395
https://leetcode.com/problems/sorting-the-sentence/discuss/1210285/2-Lines-or-Python-Solution-or-Linear-TIme-or-Linear-Memory
class Solution: def sortSentence(self, s: str) -> str: word_list = s.split() # form a list of words n = len(word_list) # total words in the list, at max 9 # dict to make k, v pairs as there are at max 9 words in the array # key as position of word, value as word without position # b...
sorting-the-sentence
2 Lines | Python Solution | Linear TIme | Linear Memory
ramanaditya
19
1,800
sorting the sentence
1,859
0.844
Easy
26,396
https://leetcode.com/problems/sorting-the-sentence/discuss/1343270/PYTHON-3-or-faster-than-95.14-or-USING-DICTIONARY
class Solution: def sortSentence(self, s: str) -> str: x = s.split() dic = {} for i in x : dic[i[-1]] = i[:-1] return ' '.join([dic[j] for j in sorted(dic)])
sorting-the-sentence
PYTHON 3 | faster than 95.14% | USING DICTIONARY
rohitkhairnar
14
933
sorting the sentence
1,859
0.844
Easy
26,397
https://leetcode.com/problems/sorting-the-sentence/discuss/1986261/Python-Easiest-Solution-With-Explanationor-99.72-Faster-or-Sorting-or-Beg-to-adv
class Solution: def sortSentence(self, s: str) -> str: splited_string = s[::-1].split() # here first we are reversing the sting and then spliting it, split() function make each word of the string as a separate element of the list. For example: ['3a', '1sihT', '4ecnetnes', '2si'] splited_st...
sorting-the-sentence
Python Easiest Solution With Explanation| 99.72% Faster | Sorting | Beg to adv
rlakshay14
3
219
sorting the sentence
1,859
0.844
Easy
26,398
https://leetcode.com/problems/sorting-the-sentence/discuss/1492510/1-line-solution-in-Python
class Solution: def sortSentence(self, s: str) -> str: return " ".join(word[:-1] for word in sorted(s.split(), key=lambda w: w[-1]))
sorting-the-sentence
1-line solution in Python
mousun224
2
292
sorting the sentence
1,859
0.844
Easy
26,399