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/complex-number-multiplication/discuss/1223014/Python-Explained-or-Easy-Solution | class Solution:
def complexNumberMultiply(self, num1: str, num2: str) -> str:
digit1 = num1.split("+") # ['1', '-1i']
digit2 = num2.split("+") # ['1', '-1i']
digit1[1] = digit1[1].replace('i','') # ['1', '-1']
digit2[1] = digit2[1].replace('i','') # ['1', '-1']
first = (int(... | complex-number-multiplication | [Python] Explained | Easy Solution | arkumari2000 | 0 | 61 | complex number multiplication | 537 | 0.714 | Medium | 9,400 |
https://leetcode.com/problems/complex-number-multiplication/discuss/872842/Python3-3-line | class Solution:
def complexNumberMultiply(self, num1: str, num2: str) -> str:
r1, i1 = map(int, num1[:-1].split('+'))
r2, i2 = map(int, num2[:-1].split('+'))
return f"{r1*r2-i1*i2}+{r1*i2+r2*i1}i" | complex-number-multiplication | [Python3] 3-line | ye15 | 0 | 39 | complex number multiplication | 537 | 0.714 | Medium | 9,401 |
https://leetcode.com/problems/complex-number-multiplication/discuss/853013/Python-3-or-Split-%2B-f-string-or-Explanation | class Solution:
def complexNumberMultiply(self, a: str, b: str) -> str:
a, b = a.split('+'), b.split('+') # split integer part and complex part
a1, b1 = int(a[0]), int(a[-1][:-1]) # convert to integer
a2, b2 = int(b[0]), int(b[-1][:-1])
x = a1*a2 # mult... | complex-number-multiplication | Python 3 | Split + f-string | Explanation | idontknoooo | 0 | 61 | complex number multiplication | 537 | 0.714 | Medium | 9,402 |
https://leetcode.com/problems/complex-number-multiplication/discuss/317795/Python-solution-32ms | class Solution:
def complexNumberMultiply(self, a: str, b: str) -> str:
k1=a.split('+')
k2=b.split('+')
(xa,ya)=(k1[0],k1[1][:-1])
(xb,yb)=(k2[0],k2[1][:-1])
return(str(int(xa)*int(xb)+int(ya)*int(yb)*-1)+'+'+str(int(xa)*int(yb)+int(ya)*int(xb))+'i') | complex-number-multiplication | Python solution 32ms | ketan35 | 0 | 63 | complex number multiplication | 537 | 0.714 | Medium | 9,403 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1057429/Python.-faster-than-100.00.-Explained-clear-and-Easy-understanding-solution.-O(n).-Recursive | class Solution:
def convertBST(self, root: TreeNode) -> TreeNode:
sum = 0
def sol(root: TreeNode) -> TreeNode:
nonlocal sum
if root:
sol(root.right)
root.val += sum
sum = root.val
sol(root.left)
return root
return sol(root) | convert-bst-to-greater-tree | Python. faster than 100.00%. Explained, clear & Easy-understanding solution. O(n). Recursive | m-d-f | 13 | 912 | convert bst to greater tree | 538 | 0.674 | Medium | 9,404 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1951325/Python3-IN-ORDER-DFS-(-*-.-*-)-Explained | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
def dfs(node, init):
if not node: return 0
r_sum = dfs(node.right, init)
orig, node.val = node.val, node.val + init + r_sum
l_sum = dfs(node.left, node.val)
... | convert-bst-to-greater-tree | ✔️ [Python3] IN-ORDER DFS ( •́ .̫ •̀ ), Explained | artod | 10 | 546 | convert bst to greater tree | 538 | 0.674 | Medium | 9,405 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1952102/Python-Simple-recursion-O(N)-%3A-Reverse-Inorder | class Solution:
def newBST(self, root, sumSoFar):
if root == None:
return sumSoFar
newSum = self.newBST(root.right, sumSoFar)
root.val = root.val + newSum
return self.newBST(root.left, root.val)
def convertBST(self, root: Optional[TreeNode]) -> ... | convert-bst-to-greater-tree | Python, Simple recursion, O(N) : Reverse Inorder | souravrane_ | 1 | 16 | convert bst to greater tree | 538 | 0.674 | Medium | 9,406 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1951959/Python-or-Java-Very-Easy-Solution | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
def traverse(root, sm):
if root is None: return sm
right = traverse(root.right, sm)
root.val += right
left = traverse(root.left, root.val)
return left
tra... | convert-bst-to-greater-tree | ✅ Python | Java Very Easy Solution | dhananjay79 | 1 | 76 | convert bst to greater tree | 538 | 0.674 | Medium | 9,407 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/2442340/Python-3-or-2-pass-solution-or-Inorder-traversal | class Solution:
def __init__(self):
self.arr = []
def getNodesInorder(self, node: Optional[TreeNode]):
if not node:
return
self.getNodesInorder(node.left)
self.arr.append(node)
self.getNodesInorder(node.right)
def convertBST(self, root: Optional[Tr... | convert-bst-to-greater-tree | Python 3 | 2-pass solution | Inorder traversal | Ploypaphat | 0 | 14 | convert bst to greater tree | 538 | 0.674 | Medium | 9,408 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1957074/Python-Intutive-solution | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
vals = []
def traverse_inorder(node):
if not node:
return
traverse_inorder(node.left)
vals.append(node.val)
traverse_inorder(node.right)
# ... | convert-bst-to-greater-tree | Python Intutive solution | pradeep288 | 0 | 20 | convert bst to greater tree | 538 | 0.674 | Medium | 9,409 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1954903/3-ways-of-traverse-and-sum-or-faster-than-98.49-or-Python3 | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
nodes = []
def inorder(root: Optional[TreeNode]):
if root != None:
inorder(root.left)
nodes.append(root)
inorder(root.right)
inorder(root)
... | convert-bst-to-greater-tree | 3 ways of traverse and sum | faster than 98.49% | Python3 | qihangtian | 0 | 5 | convert bst to greater tree | 538 | 0.674 | Medium | 9,410 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1954903/3-ways-of-traverse-and-sum-or-faster-than-98.49-or-Python3 | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
self.total = 0
def traverse(node: Optional[TreeNode]):
if node != None:
traverse(node.right)
self.total += node.val
node.val = self.total
... | convert-bst-to-greater-tree | 3 ways of traverse and sum | faster than 98.49% | Python3 | qihangtian | 0 | 5 | convert bst to greater tree | 538 | 0.674 | Medium | 9,411 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1954903/3-ways-of-traverse-and-sum-or-faster-than-98.49-or-Python3 | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
total = 0
stack = []
node = root
while node or stack:
if node:
stack.append(node)
node = node.right
else:
temp = stack.pop()
... | convert-bst-to-greater-tree | 3 ways of traverse and sum | faster than 98.49% | Python3 | qihangtian | 0 | 5 | convert bst to greater tree | 538 | 0.674 | Medium | 9,412 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1953799/Simple-and-Intuitive-Python-Solution | class Solution:
# added the param 'passdown' for easier recurrsion
def convertBST(self, root: Optional[TreeNode], passdown=0) -> Optional[TreeNode]:
if not root:
return None
root.val = self.getNewVal(root, passdown)
root.left = self.convertBST(root.left, root.val)
if... | convert-bst-to-greater-tree | Simple and Intuitive Python Solution | Mowei | 0 | 7 | convert bst to greater tree | 538 | 0.674 | Medium | 9,413 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1953029/Python-Easy-Python-Solution-Using-Array-and-Recursion | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
array = []
def InOrder(node):
if node != None:
InOrder(node.left)
array.append(node.val)
InOrder(node.right)
InOrder(root)
newnode = root
def ReplaceInOrder(newnode):
if newnode != None:
Repla... | convert-bst-to-greater-tree | [ Python ] ✅✅ Easy Python Solution Using Array and Recursion ✌🥳 | ASHOK_KUMAR_MEGHVANSHI | 0 | 12 | convert bst to greater tree | 538 | 0.674 | Medium | 9,414 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1953024/Python-Solution-or-Recursive-Inverted-Inorder-Traversal-or-Over-99-Faster-or-Simple-Logic | class Solution:
def __init__(self):
self.sum = 0
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if root is None:
return None
self.convertBST(root.right)
self.sum += root.val
root.val = self.sum
self.convertBST(root.left)
... | convert-bst-to-greater-tree | Python Solution | Recursive Inverted Inorder Traversal | Over 99% Faster | Simple Logic | Gautam_ProMax | 0 | 11 | convert bst to greater tree | 538 | 0.674 | Medium | 9,415 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1952454/Python-or-Very-Easy-To-Understand-or-Reverse-Inorder | class Solution:
def reverseInorder(self, root):
if root is None:
return None
if root.right:
root.right = self.reverseInorder(root.right)
self.sm += root.val
root.val = self.sm
if root.left:
root.left = self.reverseInorder(root.le... | convert-bst-to-greater-tree | Python | Very Easy To Understand | Reverse Inorder | sathwickreddy | 0 | 36 | convert bst to greater tree | 538 | 0.674 | Medium | 9,416 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1952283/Python-or-Reverse-Inorder-Traversal-or-Beats-98 | class Solution(object):
def convertBST(self, root):
def inorder(root, vals):
if not root: return
inorder(root.right, vals)
if not vals:
vals.append(root.val)
else:
root.val += vals[-1]
vals.append(root.val)
... | convert-bst-to-greater-tree | Python | Reverse Inorder Traversal | Beats 98% | prajyotgurav | 0 | 24 | convert bst to greater tree | 538 | 0.674 | Medium | 9,417 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1952235/~10-line-recursive-DFS-in-Python | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
def dfs(n: TreeNode = root, prev: int = 0) -> Tuple[TreeNode, int]:
if not n:
return None, 0
r, rs = dfs(n.right, prev)
val = n.val + rs + prev
l, ls = dfs... | convert-bst-to-greater-tree | ~10-line recursive DFS in Python | mousun224 | 0 | 22 | convert bst to greater tree | 538 | 0.674 | Medium | 9,418 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1951736/My-python3-solution | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if root is None:
return root
arr=[]#initializing array for prefix sum
mp={}# required map to map all the prefix sum values to values in BST
def fun(root):# doing an inorder travesal
... | convert-bst-to-greater-tree | My python3 solution | arpit92_8 | 0 | 6 | convert bst to greater tree | 538 | 0.674 | Medium | 9,419 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1951679/python-recursion | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
def convert(root, val):
if root is None:
return 0
rval = convert(root.right, val)
root.val += rval + val
lval = convert(root.left, root.val)
... | convert-bst-to-greater-tree | python recursion | user2613C | 0 | 13 | convert bst to greater tree | 538 | 0.674 | Medium | 9,420 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1951320/Simple-python-recursion-with-an-accumulator-Beats-~90 | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
def convertGreaterTree(node, acc):
if not node:
return acc
right = convertGreaterTree(node.right, acc)
node.val += right
#... | convert-bst-to-greater-tree | ✔️Simple python recursion with an accumulator - Beats ~90% | constantine786 | 0 | 20 | convert bst to greater tree | 538 | 0.674 | Medium | 9,421 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1881877/Python-easy-and-intuitive-solution-using-inorder-traversal | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if not root:
return None
stack = [root]
values = []
def inorder(node):
if node:
inorder(node.left)
values.append(node.val)
... | convert-bst-to-greater-tree | Python easy and intuitive solution using inorder traversal | byuns9334 | 0 | 24 | convert bst to greater tree | 538 | 0.674 | Medium | 9,422 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1837109/Simple-Python-Solution | class Solution:
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
def inorder(root):
elements=[]
if root is None:
return []
elements+=inorder(root.left)
elements.append(root.val)
elements+=inorder(root.r... | convert-bst-to-greater-tree | Simple Python Solution | Siddharth_singh | 0 | 23 | convert bst to greater tree | 538 | 0.674 | Medium | 9,423 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1581330/Pyhton3-Solution | class Solution:
val = 0
def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if root:
if root.right : self.convertBST(root.right)
self.val = root.val = root.val + self.val
if root.left : self.convertBST(root.left)
return root | convert-bst-to-greater-tree | Pyhton3 Solution | satyam2001 | 0 | 41 | convert bst to greater tree | 538 | 0.674 | Medium | 9,424 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1434471/Python3-Recursive-dfs-solution | class Solution:
def traversal(self, node, total):
if not node:
return None
self.traversal(node.right, total)
total[0] += node.val
node.val = total[0]
self.traversal(node.left, total)
def convertBST(self, root: Optional[TreeN... | convert-bst-to-greater-tree | [Python3] Recursive dfs solution | maosipov11 | 0 | 45 | convert bst to greater tree | 538 | 0.674 | Medium | 9,425 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1380953/Simple-Python-Recursive-Solution | class Solution:
def __init__(self):
self.sum = 0
def helper(self,root,val):
if not root:
return val
else:
returnedVal = self.helper(root.right,val)
root.val += returnedVal
return self.helper(root.left,root.val)
... | convert-bst-to-greater-tree | Simple Python Recursive Solution | Akarshippili | 0 | 48 | convert bst to greater tree | 538 | 0.674 | Medium | 9,426 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1058359/Python3-beats-99-speed-reversed-inorder-traversal | class Solution:
def convertBST(self, root: TreeNode) -> TreeNode:
if not root:
return None
def update_nodes(root: TreeNode, nodes_sum: int) -> int:
# base case, don't add to the sum
if not root:
return nodes_sum
# add to the value of current node the su... | convert-bst-to-greater-tree | [Python3] beats 99% speed, reversed inorder traversal | dbialon | 0 | 16 | convert bst to greater tree | 538 | 0.674 | Medium | 9,427 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/1057417/Python-Easy-DFS-approach-95%2B | class Solution:
def dfs(self, root: TreeNode, s) -> int:
if root.right:
s = self.dfs(root.right, s)
root.val += s
s = root.val
if root.left:
s = self.dfs(root.left, s)
return s
def convertBST(self, root: TreeNode) -> TreeNode:
if root:
self.dfs(root, 0)
return root | convert-bst-to-greater-tree | Python Easy DFS approach 95%+ | SlavaHerasymov | 0 | 34 | convert bst to greater tree | 538 | 0.674 | Medium | 9,428 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/872828/Python3-inorder-traversal | class Solution:
def convertBST(self, root: TreeNode) -> TreeNode:
def fn(node, x):
"""Inorder traverse the tree and update node's value."""
if not node: return x
x = fn(node.right, x) # sum of right subtree
x += node.val
node.val = x
... | convert-bst-to-greater-tree | [Python3] inorder traversal | ye15 | 0 | 45 | convert bst to greater tree | 538 | 0.674 | Medium | 9,429 |
https://leetcode.com/problems/convert-bst-to-greater-tree/discuss/872828/Python3-inorder-traversal | class Solution:
def convertBST(self, root: TreeNode) -> TreeNode:
val = 0
node, stack = root, []
while stack or node:
if node:
stack.append(node)
node = node.right
else:
node = stack.pop()
node.val = v... | convert-bst-to-greater-tree | [Python3] inorder traversal | ye15 | 0 | 45 | convert bst to greater tree | 538 | 0.674 | Medium | 9,430 |
https://leetcode.com/problems/minimum-time-difference/discuss/1829297/python-3-bucket-sort-O(n)-time-O(1)-space | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
M = 1440
times = [False] * M
for time in timePoints:
minute = self.minute(time)
if times[minute]:
return 0
times[minute] = True
minutes = [i for i i... | minimum-time-difference | [python 3] bucket sort, O(n) time, O(1) space | dereky4 | 9 | 595 | minimum time difference | 539 | 0.563 | Medium | 9,431 |
https://leetcode.com/problems/minimum-time-difference/discuss/1995265/Clean-Python-O(n)-vs-O(n-log-n)-Discussion-Shortcut-for-long-inputs | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
def toMinFromZero(t):
return 60*int(t[:2]) + int(t[3:])
maxMins = 60*24 # number of mins in a day
if len(timePoints) > maxMins:
return 0
# we sort the timestamps. Idea: Th... | minimum-time-difference | Clean Python, O(n) vs O(n log n) Discussion, Shortcut for long inputs | boris17 | 8 | 528 | minimum time difference | 539 | 0.563 | Medium | 9,432 |
https://leetcode.com/problems/minimum-time-difference/discuss/488280/Python-Simple-Solution-2-Liner | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
minutes = sorted(list(map(lambda x: int(x[:2]) * 60 + int(x[3:]), timePoints)))
return min((y - x) % (24 * 60) for x, y in zip(minutes, minutes[1:] + minutes[:1])) | minimum-time-difference | Python - Simple Solution - 2 Liner | mmbhatk | 4 | 951 | minimum time difference | 539 | 0.563 | Medium | 9,433 |
https://leetcode.com/problems/minimum-time-difference/discuss/1676798/WEEB-DOES-PYTHON | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
timePoints.sort()
def getTimeDiff(timeString1, timeString2):
time1 = int(timeString1[:2]) * 60 + int(timeString1[3:])
time2 = int(timeString2[:2]) * 60 + int(timeString2[3:])
minDiff = abs(time1 - time2)
return min(minDiff, 1... | minimum-time-difference | WEEB DOES PYTHON | Skywalker5423 | 3 | 149 | minimum time difference | 539 | 0.563 | Medium | 9,434 |
https://leetcode.com/problems/minimum-time-difference/discuss/846858/Python-3-or-Sort-Math-or-Explanations | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
def minute(time_str):
h, m = time_str.split(':')
return int(h)*60+int(m)
total, minutes = 24*60, sorted([minute(time_str) for time_str in timePoints])
n, diff = len(minutes), total - (minutes[-... | minimum-time-difference | Python 3 | Sort, Math | Explanations | idontknoooo | 2 | 560 | minimum time difference | 539 | 0.563 | Medium | 9,435 |
https://leetcode.com/problems/minimum-time-difference/discuss/2438080/Python-Solution-with-Sorting-and-Two-pointers | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
minutesConvert = []
# convert time points to minutes expression
for time in timePoints:
t = time.split(":")
minutes = int(t[0]) * 60 + int(t[1])
minutesConvert.append(minutes)
# sort th... | minimum-time-difference | Python Solution with Sorting and Two-pointers | siyu_ | 1 | 152 | minimum time difference | 539 | 0.563 | Medium | 9,436 |
https://leetcode.com/problems/minimum-time-difference/discuss/872912/Python3-sort-the-time | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
def fn(t):
"""Return time point as minute."""
h, m = map(int, t.split(":"))
return 60*h + m
timePoints = sorted(map(fn, timePoints))
timePoints += [timePoints[0]... | minimum-time-difference | [Python3] sort the time | ye15 | 1 | 141 | minimum time difference | 539 | 0.563 | Medium | 9,437 |
https://leetcode.com/problems/minimum-time-difference/discuss/872912/Python3-sort-the-time | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
timePoints = sorted(60*int(x[:2]) + int(x[3:]) for x in timePoints) # sorting
timePoints += [timePoints[0]]
return min((timePoints[i] - timePoints[i-1])%1440 for i in range(1, len(timePoints))) | minimum-time-difference | [Python3] sort the time | ye15 | 1 | 141 | minimum time difference | 539 | 0.563 | Medium | 9,438 |
https://leetcode.com/problems/minimum-time-difference/discuss/710624/Python3-sort-based-solution-Minimum-Time-Difference | class Solution:
def findMinDifference(self, time: List[str]) -> int:
def minute(a):
return int(a[:2]) * 60 + int(a[-2:])
time.sort()
ans = 12 * 60
for i in range(len(time)):
j = (i + 1) % len(time)
ans = min(ans, (minute(time[j]) - minute(time[i]))... | minimum-time-difference | Python3 sort-based solution - Minimum Time Difference | r0bertz | 1 | 327 | minimum time difference | 539 | 0.563 | Medium | 9,439 |
https://leetcode.com/problems/minimum-time-difference/discuss/2829819/SIMPLE-PYTHON-SOLUTION | class Solution:
def toint(s):
h=int(s[0:2])
m=int(s[3:5])
return int(h*60+m)
def findMinDifference(self, t: List[str]) -> int:
b=set(t)
if(len(b)<len(t)):
return 0
def toint(s):
h=int(s[0:2])
m=int(s[3:5])
retu... | minimum-time-difference | SIMPLE PYTHON SOLUTION | _gajera__28_0 | 0 | 5 | minimum time difference | 539 | 0.563 | Medium | 9,440 |
https://leetcode.com/problems/minimum-time-difference/discuss/2597155/Python-Concise-but-readable-Solution | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
# get the minutes
timePoints = list(map(self.get_minutes, timePoints))
# sort the time points
timePoints.sort()
# append the earlies value increased by one day (to get differ... | minimum-time-difference | [Python] - Concise but readable Solution | Lucew | 0 | 32 | minimum time difference | 539 | 0.563 | Medium | 9,441 |
https://leetcode.com/problems/minimum-time-difference/discuss/2366803/Python-solution-(easy-to-understand) | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
timePoints.sort()
minDiff = float("inf")
for i in range(1, len(timePoints)):
prev = timePoints[i - 1]
curr = timePoints[i]
if curr == prev:
return 0
prev... | minimum-time-difference | Python solution (easy to understand) | lau125 | 0 | 43 | minimum time difference | 539 | 0.563 | Medium | 9,442 |
https://leetcode.com/problems/minimum-time-difference/discuss/2061993/Python-sorting | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
minutes = sorted(60*int(t[:2]) + int(t[3:]) for t in timePoints)
diff = minutes[0] - minutes[-1] + 24 * 60
for i in range(len(minutes)-1):
diff = min(diff, minutes[i+1] - minutes[i])
... | minimum-time-difference | Python, sorting | blue_sky5 | 0 | 132 | minimum time difference | 539 | 0.563 | Medium | 9,443 |
https://leetcode.com/problems/minimum-time-difference/discuss/2039125/fast-python | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
times = [
(int(time.split(":")[0])*60 + int(time.split(":")[1]))
for time in timePoints]
times = sorted(times)
for t in times.copy():
if t < 12*60:
times.ap... | minimum-time-difference | fast python | hermesdt | 0 | 108 | minimum time difference | 539 | 0.563 | Medium | 9,444 |
https://leetcode.com/problems/minimum-time-difference/discuss/1735026/python3-Solution-with-using-sorting | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
res = float('inf')
timePoints.sort()
timePoints[0] = int(timePoints[0][:2]) * 60 + int(timePoints[0][3:])
for idx in range(1, len(timePoints)):
timePoints[idx] = int(time... | minimum-time-difference | [python3] Solution with using sorting | maosipov11 | 0 | 62 | minimum time difference | 539 | 0.563 | Medium | 9,445 |
https://leetcode.com/problems/minimum-time-difference/discuss/1321140/Beginner-Friendly | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
noDuplicates = set(timePoints)
if len(noDuplicates) != len(timePoints):
return 0
timePoints = [time.split(':') for time in timePoints]
print(timePoints)
timePoints.sort(key = lambda x:(int(... | minimum-time-difference | Beginner Friendly | WinstonWolfe07 | 0 | 163 | minimum time difference | 539 | 0.563 | Medium | 9,446 |
https://leetcode.com/problems/minimum-time-difference/discuss/908148/Python-easy-solution | class Solution:
def findMinDifference(self, timePoints: List[str]) -> int:
temp = []
for t in timePoints:
hour, minute = t.split(":")
temp.append(int(hour)*60+int(minute))
# additional check to count more than 24h
... | minimum-time-difference | Python easy solution | ermolushka2 | 0 | 167 | minimum time difference | 539 | 0.563 | Medium | 9,447 |
https://leetcode.com/problems/minimum-time-difference/discuss/413812/Python3-88-runtime-%2B-100-memory | class Solution:
def time2min(self, t):
return int(t[:2]) * 60 + int(t[3:])
def findMinDifference(self, timePoints: List[str]) -> int:
new = [self.time2min(t) for t in timePoints]
new.sort()
offset = 60*24
new.extend([n+offset for n in new])
... | minimum-time-difference | Python3 88% runtime + 100% memory | bear_sun | 0 | 368 | minimum time difference | 539 | 0.563 | Medium | 9,448 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
counts = defaultdict(int)
for num in nums:
counts[num] += 1
for num, count in counts.items():
if count == 1:
return num
return -1 # this will never be reached
# return Cou... | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,449 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
val, seen = -1, True
for num in nums:
if val == num:
seen = True
elif seen:
val = num
seen = False
else:
return val
return... | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,450 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
for i in range(0, len(nums)-1, 2): # pairwise comparison
if nums[i] != nums[i+1]: # found the single element
return nums[i]
return nums[-1] # the last element is the single element | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,451 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
result = 0
for i in range(len(nums)):
if i%2: # alternate between +ve and -ve
result -= nums[i]
else:
result += nums[i]
return result
# return sum((-1)**i*v for i,... | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,452 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
result = 0
for num in nums:
result ^= num
return result
# return reduce(xor, nums) # one-liner | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,453 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
result = nums[0]
for i in range(1, len(nums), 2):
result += nums[i+1]-nums[i]
return result
# return nums[0] + sum(nums[i+1]-nums[i] for i in range(1, len(nums), 2)) # one-liner | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,454 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
result = nums[0]
for i in range(1, len(nums), 2):
result ^= nums[i]^nums[i+1]
return result
# return reduce(lambda x,i: x^nums[i]^nums[i+1], range(1, len(nums), 2), nums[0]) # one-liner | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,455 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
lo, hi = 0, len(nums)-1
while lo < hi:
mid = lo+(hi-lo)//2
if nums[mid] == nums[mid-1]: # duplicate found
if mid%2: # target > mid
lo = mid+1 # exclude second ind... | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,456 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587293/Python-3-Simple-Approaches-with-Explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
lo, hi = 0, len(nums)-2 # hi starts from an even index so that hi^1 gives the next odd number
while lo <= hi:
mid = lo+(hi-lo)//2
if nums[mid] == nums[mid^1]:
lo = mid+1
else:
... | single-element-in-a-sorted-array | [Python] 3 Simple Approaches with Explanation | zayne-siew | 54 | 2,800 | single element in a sorted array | 540 | 0.585 | Medium | 9,457 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/627989/Python-O(lg-n)-by-binary-search-85%2B-w-Visualization | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
size = len(nums)
left, right = 0, size // 2
while left < right:
pair_index = left + ( right - left ) // 2
if nums[2*pair_index] != nums[2*pair_in... | single-element-in-a-sorted-array | Python O(lg n) by binary search 85%+ [w/ Visualization] | brianchiang_tw | 7 | 714 | single element in a sorted array | 540 | 0.585 | Medium | 9,458 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1588059/Simple-One-Liner-or-Python | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
return Counter(nums).most_common()[-1][0] | single-element-in-a-sorted-array | Simple One Liner | Python | deep765 | 3 | 109 | single element in a sorted array | 540 | 0.585 | Medium | 9,459 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1588464/Simple-Python-Solution | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
c=0
for i in nums:
if nums.count(i)==1:
return i | single-element-in-a-sorted-array | Simple Python Solution | FaizanAhmed_ | 2 | 147 | single element in a sorted array | 540 | 0.585 | Medium | 9,460 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2833161/defaultdict-does-the-job | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
dc=defaultdict(lambda:0)
for a in nums:
dc[a]+=1
for a in dc:
if(dc[a]==1):
return a | single-element-in-a-sorted-array | defaultdict does the job | droj | 0 | 1 | single element in a sorted array | 540 | 0.585 | Medium | 9,461 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2829267/Binary-search-on-two-conditions | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
left, right = 0, len(nums) - 1
while left <= right:
mid = left + (right - left) // 2
if mid > 0 and nums[mid] == nums[mid - 1]:
if (mid - 1) % 2 == 0:
left = mid + 1
... | single-element-in-a-sorted-array | Binary search on two conditions | michaelniki | 0 | 4 | single element in a sorted array | 540 | 0.585 | Medium | 9,462 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2810156/Simple-Python-Solution-%3A-Time-Complexity-%3A-O(LogN)-or-Space-Complexity-%3A-O(1) | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
start = 0
end = len(nums)-1
while start <= end:
if start == end:
return nums[start]
mid = (start+end) // 2
if nums[mid] != nums[mid+1] and nums[mid]!= nums[mi... | single-element-in-a-sorted-array | Simple Python Solution : Time Complexity : O(LogN) | Space Complexity : O(1) | MaviOp | 0 | 3 | single element in a sorted array | 540 | 0.585 | Medium | 9,463 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2804580/Simple-Python-SolutionororBinary-SearchororO(log2n) | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
n=len(nums)
low=0
high=n-1
while(low <= high):
mid = (low+high)//2
if(( mid==n-1 or nums[mid] != nums[mid+1] ) and ( mid==0 or nums[mid]!=nums[mid-1])):
return(nums[mid])
... | single-element-in-a-sorted-array | Simple Python Solution||Binary Search||O(log2n) | Ankit_Verma03 | 0 | 1 | single element in a sorted array | 540 | 0.585 | Medium | 9,464 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2802481/Python-Easy-to-understand-using-lambda | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
d = {}
for word in nums:
d[word] = d.get(word, 0) + 1
#print(d)
res = sorted(d.keys(), key=lambda word: d[word])
#print(res)
return res[0] | single-element-in-a-sorted-array | Python Easy to understand, using lambda | subbhashitmukherjee | 0 | 2 | single element in a sorted array | 540 | 0.585 | Medium | 9,465 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2788513/Simple-python-binary-search-explanation | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
lo, hi = 0, len(nums) - 1
while lo < hi:
mid = (lo + hi) // 2
if mid % 2 == 0:
# make sure the number of elements between index 0 and index mid (inclusive) is always is even
... | single-element-in-a-sorted-array | Simple python binary search explanation | lister777 | 0 | 6 | single element in a sorted array | 540 | 0.585 | Medium | 9,466 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2753181/using-dictionary-python-3 | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
dict={}
for i in nums:
if i not in dict:
dict[i]=1
else:
dict[i]+=1
for x,y in dict.items():
if y==1:
return x | single-element-in-a-sorted-array | using dictionary python-3 | Kevin7777777 | 0 | 2 | single element in a sorted array | 540 | 0.585 | Medium | 9,467 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2445714/Binary-search-problem-practise | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
low = 0
high = len(nums)-1
while(low<=high):
mid = (low+high)//2
if(low == high):
return nums[low]
if(mid%2 == 0):
if(nums[mid] == n... | single-element-in-a-sorted-array | Binary search problem practise | sandhya_vollala | 0 | 36 | single element in a sorted array | 540 | 0.585 | Medium | 9,468 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2372119/Single-Element-in-a-Sorted-Array | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
count=collections.Counter(nums)
for k,v in count.items():
if v==1:return k | single-element-in-a-sorted-array | Single Element in a Sorted Array | Faraz369 | 0 | 13 | single element in a sorted array | 540 | 0.585 | Medium | 9,469 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2346863/C%2B%2BPython-O(N)-to-O(logN)-Solution | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
res = 0
for i in range(len(nums)):
res = res^nums[i]
return res | single-element-in-a-sorted-array | C++/Python O(N) to O(logN) Solution | arpit3043 | 0 | 57 | single element in a sorted array | 540 | 0.585 | Medium | 9,470 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2346863/C%2B%2BPython-O(N)-to-O(logN)-Solution | class Solution(object):
def singleNonDuplicate(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
left, right = 0, len(nums) - 1
while left + 1 < right:
mid = (left + right) // 2
if mid % 2 == 1:
if nums[mid] == nums[mid - 1... | single-element-in-a-sorted-array | C++/Python O(N) to O(logN) Solution | arpit3043 | 0 | 57 | single element in a sorted array | 540 | 0.585 | Medium | 9,471 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2284683/Python3-Binary-search | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
left, right = 0, len(nums) - 1
if right == 0:
return nums[right]
if nums[0]!=nums[1]:
return nums[0]
if nums[right] != nums[right-1]:
return nums[right]
while(left<=right... | single-element-in-a-sorted-array | Python3 Binary search | yashchandani98 | 0 | 19 | single element in a sorted array | 540 | 0.585 | Medium | 9,472 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2219612/Python-Solution-or-2-Approaches-or-TC%3A-O(n)-and-O(log(n)) | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
# 1st Approach
# TC: O(n)
# xor=0
# for i in range(len(nums)):
# xor^=nums[i]
# return xor
# 2nd Approach
# using binary search
# TC: O(log(n))... | single-element-in-a-sorted-array | Python Solution | 2 Approaches | TC:- O(n) and O(log(n)) | Siddharth_singh | 0 | 54 | single element in a sorted array | 540 | 0.585 | Medium | 9,473 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2192157/Easy-Approach | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
n = len(nums)
i = 1
if n == 1:
return nums[0]
while i < n:
if nums[i] != nums[i - 1]:
return nums[i - 1]
i += 2
return nums[i - 1] | single-element-in-a-sorted-array | Easy Approach | Vaibhav7860 | 0 | 50 | single element in a sorted array | 540 | 0.585 | Medium | 9,474 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2110027/Python-or-Two-Different-Solution-using-Binary-Search-and-hashmap | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
# ///////// Solution 2 TC: O(log N) /////////
l,r = 0, len(nums) - 1
while l < r:
mid = (l+r)//2
mid = mid - 1 if mid % 2 == 1 else mid
if nums[mid] == nums[mid+1]:
... | single-element-in-a-sorted-array | Python | Two Different Solution using Binary Search and hashmap | __Asrar | 0 | 66 | single element in a sorted array | 540 | 0.585 | Medium | 9,475 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/2091061/Python-1-Liner-Solution | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
return reduce(operator.ixor, nums) | single-element-in-a-sorted-array | Python 1-Liner Solution | VictorSG | 0 | 39 | single element in a sorted array | 540 | 0.585 | Medium | 9,476 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1922641/2-Python-Solutions | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
return Counter(nums).most_common()[-1][0] | single-element-in-a-sorted-array | 2 Python Solutions | Taha-C | 0 | 33 | single element in a sorted array | 540 | 0.585 | Medium | 9,477 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1922641/2-Python-Solutions | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
C=defaultdict(int)
for num in nums: C[num]+=1
for num,c in C.items():
if c==1: return num | single-element-in-a-sorted-array | 2 Python Solutions | Taha-C | 0 | 33 | single element in a sorted array | 540 | 0.585 | Medium | 9,478 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1922641/2-Python-Solutions | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
nums.append(100001)
for i in range(0,len(nums)-1,2):
if nums[i]!=nums[i+1]: return nums[i] | single-element-in-a-sorted-array | 2 Python Solutions | Taha-C | 0 | 33 | single element in a sorted array | 540 | 0.585 | Medium | 9,479 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1922641/2-Python-Solutions | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
return sum((-1)**i*v for i,v in enumerate(nums)) | single-element-in-a-sorted-array | 2 Python Solutions | Taha-C | 0 | 33 | single element in a sorted array | 540 | 0.585 | Medium | 9,480 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1922641/2-Python-Solutions | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
lo=0 ; hi=len(nums)-1
while lo<hi:
mid=2*((lo+hi)//4)
if nums[mid]==nums[mid+1]: lo=mid+2
else: hi=mid
return nums[lo] | single-element-in-a-sorted-array | 2 Python Solutions | Taha-C | 0 | 33 | single element in a sorted array | 540 | 0.585 | Medium | 9,481 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1808800/Python-or-Xor-or-simple-or-basic | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
val=0
for i in range(len(nums)):
val=val^nums[i]
return val | single-element-in-a-sorted-array | Python | Xor | simple | basic | InvincibleTaki | 0 | 30 | single element in a sorted array | 540 | 0.585 | Medium | 9,482 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1588736/100-fastest-log(n)-solution-with-clear-comments | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
# noting the low and high
l, h = 0, len(nums) - 1
# boundary conditions
if h == 0:
return nums[0]
elif nums[l] != nums[l+1]:
return nums[l]
elif nums[h] != nums[h-1]:
... | single-element-in-a-sorted-array | 100% fastest - log(n) solution with clear comments | PSC_Crack | 0 | 74 | single element in a sorted array | 540 | 0.585 | Medium | 9,483 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1588686/Python-3-binary-search-faster-than-99 | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
n = len(nums) - 1
left, right = 0, n
while left <= right:
mid = (left + right) // 2
if (mid == 0 or nums[mid-1] != nums[mid]) and (mid == n or nums[mid+1] != nums[mid]):
return nums... | single-element-in-a-sorted-array | Python 3 binary search faster than 99% | dereky4 | 0 | 116 | single element in a sorted array | 540 | 0.585 | Medium | 9,484 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587845/One-pass-with-step-2-99-speed | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
for i in range(0, len(nums) - 1, 2):
if nums[i] != nums[i + 1]:
return nums[i]
return nums[-1] | single-element-in-a-sorted-array | One pass with step 2, 99% speed | EvgenySH | 0 | 20 | single element in a sorted array | 540 | 0.585 | Medium | 9,485 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1587418/Python3-simple-soln.-0-sliding-window | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
dictN = {} # hashtable to store element reptition count
if len(nums) == 1: #boundary condition . if length of nums is 1 then return 1
return 1
for j,i in enumerate(nums):
if i not in dictN:
dictN[i] = 0 #initiate count to 0
... | single-element-in-a-sorted-array | Python3 simple soln. 0 sliding window | golden-eagle | 0 | 16 | single element in a sorted array | 540 | 0.585 | Medium | 9,486 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1582712/Python-oror-Easy-Solution-oror-O(logn)-and-O(1) | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
if len(nums) == 1:
return nums[0]
l, r = 0, len(nums) - 1
while l < r:
mid = l + (r - l) // 2
if nums[mid] == nums[mid + 1]:
if (r - mid) % 2 == 0:
l = mid + 2
else:
r = mid - 1
elif nums[mid - 1] == nums[mid]... | single-element-in-a-sorted-array | Python || Easy Solution || O(logn) and O(1) | naveenrathore | 0 | 99 | single element in a sorted array | 540 | 0.585 | Medium | 9,487 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1578323/Py3Py-Very-simple-solution-using-two-pointers-w-comments | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
# Init
i = 0
j = len(nums) - 1
# Update till unique number not found
while i < j:
# Increment i
if nums[i] == nums[i+1]:
i = i+2
... | single-element-in-a-sorted-array | [Py3/Py] Very simple solution using two pointers w/ comments | ssshukla26 | 0 | 44 | single element in a sorted array | 540 | 0.585 | Medium | 9,488 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1559678/Python3-Solution-with-using-binary-search | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) // 2
isOdd = (right - mid) % 2 != 0 # isOdd is flag that there is an odd number of elements in the left part relative to the ... | single-element-in-a-sorted-array | [Python3] Solution with using binary search | maosipov11 | 0 | 51 | single element in a sorted array | 540 | 0.585 | Medium | 9,489 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1456184/Simple-Python-O(logn)-binary-search-%2B-parity-check-solution | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
# monotonicity is a step function:
# ___________ yes single
# |
# no single _______|
# the problem boils down to determining whether
# the single number exists to ... | single-element-in-a-sorted-array | Simple Python O(logn) binary search + parity check solution | Charlesl0129 | 0 | 96 | single element in a sorted array | 540 | 0.585 | Medium | 9,490 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1403185/Python-logn-solution-or-Easy-to-understand-or-SR | class Solution:
def check(self, li, idx, n):
if idx-1 >= 0 and li[idx-1] == li[idx]:
return -1
elif idx+1 < n and li[idx+1] == li[idx]:
return 1
else:
return 0 #no left and no right same value
def singleNonDuplicate(self, nums: List[int]) -> ... | single-element-in-a-sorted-array | Python logn solution | Easy to understand | SR | sathwickreddy | 0 | 94 | single element in a sorted array | 540 | 0.585 | Medium | 9,491 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/1093564/Simple-Python3-Binary-search-solution | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
while 1 == 1:
len_nums = len(nums)
offset = 0
if len_nums == 1:
return nums[0]
mid = round(len_nums / 2)
if nums[mid ... | single-element-in-a-sorted-array | Simple Python3 Binary search solution | olivierkessler | 0 | 42 | single element in a sorted array | 540 | 0.585 | Medium | 9,492 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/873912/Python3-binary-search | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = lo + hi >> 1
if (mid == 0 or nums[mid - 1] != nums[mid]) and (mid+1 == len(nums) or nums[mid] != nums[mid+1]): return nums[mid]
if nums[mid] == nums[m... | single-element-in-a-sorted-array | [Python3] binary search | ye15 | 0 | 69 | single element in a sorted array | 540 | 0.585 | Medium | 9,493 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/873912/Python3-binary-search | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
lo, hi = 0, len(nums)
while lo < hi:
mid = lo + hi >> 1
if mid^1 < len(nums) and nums[mid] == nums[mid^1]: lo = mid+1
else: hi = mid
return nums[lo] | single-element-in-a-sorted-array | [Python3] binary search | ye15 | 0 | 69 | single element in a sorted array | 540 | 0.585 | Medium | 9,494 |
https://leetcode.com/problems/single-element-in-a-sorted-array/discuss/629392/Python-Recursive-solution-with-O(logn)-time-complexity-and-O(1)-space-complexity | class Solution:
def singleNonDuplicate(self, nums: List[int]) -> int:
return self.helper(nums)
def helper(self, nums):
if len(nums)==1:
return nums[0]
mid = len(nums)//2
if nums[mid]!=nums[mid-1] and nums[mid]!=nums[mid+1]:
return nums[mid]
el... | single-element-in-a-sorted-array | Python Recursive solution with O(logn) time complexity and O(1) space complexity | user8847C | 0 | 28 | single element in a sorted array | 540 | 0.585 | Medium | 9,495 |
https://leetcode.com/problems/reverse-string-ii/discuss/343424/Python-3-solution-using-recursion-(efficient)-3-liner-with-explanation | class Solution:
def reverseStr(self, s: str, k: int) -> str:
if len(s)<(k):return s[::-1]
if len(s)<(2*k):return (s[:k][::-1]+s[k:])
return s[:k][::-1]+s[k:2*k]+self.reverseStr(s[2*k:],k) | reverse-string-ii | Python 3 solution using recursion (efficient) 3-liner with explanation | ketan35 | 22 | 1,700 | reverse string ii | 541 | 0.505 | Easy | 9,496 |
https://leetcode.com/problems/reverse-string-ii/discuss/1017551/Simple-and-easy-faster-than-99.52 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
s=list(s)
for i in range(0,len(s),2*k):
if len(s[i:i+k])<k:
s[i:i+k]=s[i:i+k][::-1]
elif 2*k>len(s[i:i+k])>=k:
s[i:i+k]=s[i:i+k][::-1]
return "".join(s) | reverse-string-ii | Simple and easy - faster than 99.52% | thisisakshat | 5 | 598 | reverse string ii | 541 | 0.505 | Easy | 9,497 |
https://leetcode.com/problems/reverse-string-ii/discuss/1977347/Python3-94-faster-with-explanation | class Solution:
def reverseStr(self, s: str, k: int) -> str:
news, ogLen = '', len(s)
i = 0
while (i < ogLen) and s:
if i % 2 == 0:
news += s[:k][::-1]
else:
news += s[:k]
s = s[k:]
i += 1
return news | reverse-string-ii | Python3, 94% faster with explanation | cvelazquez322 | 2 | 302 | reverse string ii | 541 | 0.505 | Easy | 9,498 |
https://leetcode.com/problems/reverse-string-ii/discuss/1477226/Simple-or-Python-3-or-24-ms-faster-than-97.61 | class Solution:
def reverseStr(self, s: str, k: int) -> str:
r = ''
for i in range(0, len(s), k*2):
r += s[i:i+k][::-1] + s[i+k:i+k+k]
return r | reverse-string-ii | Simple | Python 3 | 24 ms, faster than 97.61% | deep765 | 2 | 205 | reverse string ii | 541 | 0.505 | Easy | 9,499 |
Subsets and Splits
Top 2 Solutions by Upvotes
Identifies the top 2 highest upvoted Python solutions for each problem, providing insight into popular approaches.