problem_title stringlengths 3 77 | python_solutions stringlengths 81 8.45k | post_href stringlengths 64 213 | upvotes int64 0 1.2k | question stringlengths 0 3.6k | post_title stringlengths 2 100 | views int64 1 60.9k | slug stringlengths 3 77 | acceptance float64 0.14 0.91 | user stringlengths 3 26 | difficulty stringclasses 3
values | __index_level_0__ int64 0 34k | number int64 1 2.48k |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
shortest palindrome | class Solution:
def shortestPalindrome(self, s: str) -> str:
end = 0
# if the string itself is a palindrome return it
if(s == s[::-1]):
return s
# Otherwise find the end index of the longest palindrome that starts
# from the first charac... | https://leetcode.com/problems/shortest-palindrome/discuss/2157861/No-DP-No-DS-Intuitive-with-comments-oror-Python | 2 | You are given a string s. You can convert s to a
palindrome
by adding characters in front of it.
Return the shortest palindrome you can find by performing this transformation.
Example 1:
Input: s = "aacecaaa"
Output: "aaacecaaa"
Example 2:
Input: s = "abcd"
Output: "dcbabcd"
Constraints:
0 <= s.length <= 5 * 104
s ... | No DP; No DS; Intuitive with comments || Python | 124 | shortest-palindrome | 0.322 | a-myth | Hard | 3,694 | 214 |
kth largest element in an array | class Solution:
def findKthLargest(self, nums: List[int], k: int) -> int:
n = len(nums)
def partition(l, r, pivot):
pivot_elem=nums[pivot]
nums[r],nums[pivot]=nums[pivot],nums[r]
index=l
for i in range(l, r):
... | https://leetcode.com/problems/kth-largest-element-in-an-array/discuss/2180509/Python-Easy-O(logn)-Space-approach-or-One-liner | 24 | Given an integer array nums and an integer k, return the kth largest element in the array.
Note that it is the kth largest element in the sorted order, not the kth distinct element.
Can you solve it without sorting?
Example 1:
Input: nums = [3,2,1,5,6,4], k = 2
Output: 5
Example 2:
Input: nums = [3,2,3,1,2,4,5,5,6], ... | Python Easy O(logn) Space approach | One liner | 3,300 | kth-largest-element-in-an-array | 0.658 | constantine786 | Medium | 3,711 | 215 |
combination sum iii | class Solution:
def combinationSum3(self, k: int, n: int) -> List[List[int]]:
current_combination, combinations = [], []
integer, combination_sum = 1, 0
queue = [(integer, current_combination, combination_sum)]
while queue:
integer, current_combination, combinati... | https://leetcode.com/problems/combination-sum-iii/discuss/1312030/Elegant-Python-Iterative-or-Recursive | 5 | Find all valid combinations of k numbers that sum up to n such that the following conditions are true:
Only numbers 1 through 9 are used.
Each number is used at most once.
Return a list of all possible valid combinations. The list must not contain the same combination twice, and the combinations may be returned in any ... | Elegant Python Iterative | Recursive | 175 | combination-sum-iii | 0.672 | soma28 | Medium | 3,761 | 216 |
contains duplicate | class Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
return len(set(nums)) != len(nums) | https://leetcode.com/problems/contains-duplicate/discuss/1496268/Python-98-speed-faster | 11 | Given an integer array nums, return true if any value appears at least twice in the array, and return false if every element is distinct.
Example 1:
Input: nums = [1,2,3,1]
Output: true
Example 2:
Input: nums = [1,2,3,4]
Output: false
Example 3:
Input: nums = [1,1,1,3,3,4,3,2,4,2]
Output: true
Constraints:
1 <= num... | Python // 98% speed faster | 1,900 | contains-duplicate | 0.613 | fabioo29 | Easy | 3,811 | 217 |
the skyline problem | class Solution:
def getSkyline(self, buildings: List[List[int]]) -> List[List[int]]:
# for the same x, (x, -H) should be in front of (x, 0)
# For Example 2, we should process (2, -3) then (2, 0), as there's no height change
x_height_right_tuples = sorted([(L, -H, R) for L, R, H in buildings]... | https://leetcode.com/problems/the-skyline-problem/discuss/2640697/Python-oror-Easily-Understood-oror-Faster-oror-with-maximum-heap-explained | 12 | A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Given the locations and heights of all the buildings, return the skyline formed by these buildings collectively.
The geometric information of each building is given in the array buildings where ... | 🔥 Python || Easily Understood ✅ || Faster || with maximum heap explained | 1,100 | the-skyline-problem | 0.416 | rajukommula | Hard | 3,856 | 218 |
contains duplicate ii | class Solution:
def containsNearbyDuplicate(self, nums: List[int], k: int) -> bool:
# Create hset for storing previous of k elements...
hset = {}
# Traverse for all elements of the given array in a for loop...
for idx in range(len(nums)):
# If duplicate element is present... | https://leetcode.com/problems/contains-duplicate-ii/discuss/2463150/Very-Easy-oror-100-oror-Fully-Explained-oror-Java-C%2B%2B-Python-Javascript-Python3-(Using-HashSet) | 45 | Given an integer array nums and an integer k, return true if there are two distinct indices i and j in the array such that nums[i] == nums[j] and abs(i - j) <= k.
Example 1:
Input: nums = [1,2,3,1], k = 3
Output: true
Example 2:
Input: nums = [1,0,1,1], k = 1
Output: true
Example 3:
Input: nums = [1,2,3,1,2,3], k = 2... | Very Easy || 100% || Fully Explained || Java, C++, Python, Javascript, Python3 (Using HashSet) | 3,400 | contains-duplicate-ii | 0.423 | PratikSen07 | Easy | 3,870 | 219 |
contains duplicate iii | class Solution:
def containsDuplicate(self, nums: List[int]) -> bool:
seen = set()
for x in nums:
if x in seen: return True
seen.add(x)
return False | https://leetcode.com/problems/contains-duplicate-iii/discuss/825267/Python3-summarizing-Contain-Duplicates-I-II-III | 22 | You are given an integer array nums and two integers indexDiff and valueDiff.
Find a pair of indices (i, j) such that:
i != j,
abs(i - j) <= indexDiff.
abs(nums[i] - nums[j]) <= valueDiff, and
Return true if such pair exists or false otherwise.
Example 1:
Input: nums = [1,2,3,1], indexDiff = 3, valueDiff = 0
Output: ... | [Python3] summarizing Contain Duplicates I, II, III | 689 | contains-duplicate-iii | 0.22 | ye15 | Hard | 3,921 | 220 |
maximal square | class Solution:
def maximalSquare(self, matrix: List[List[str]]) -> int:
result = 0
for i in range(len(matrix)):
for j in range(len(matrix[0])):
curr = 0 # current length of the square at (i, j)
flag = True # indicates if there still exists a valid squar... | https://leetcode.com/problems/maximal-square/discuss/1632285/Python-1D-Array-DP-Optimisation-Process-Explained | 9 | Given an m x n binary matrix filled with 0's and 1's, find the largest square containing only 1's and return its area.
Example 1:
Input: matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]]
Output: 4
Example 2:
Input: matrix = [["0","1"],["1","0"]]
Output: 1
Example 3:
In... | [Python] 1D-Array DP - Optimisation Process Explained | 610 | maximal-square | 0.446 | zayne-siew | Medium | 3,932 | 221 |
count complete tree nodes | class Solution:
def countNodes(self, root: TreeNode) -> int:
if not root: return 0
return 1 + self.countNodes(root.left) + self.countNodes(root.right) | https://leetcode.com/problems/count-complete-tree-nodes/discuss/701577/PythonPython3-Count-Complete-Tree-Nodes | 4 | Given the root of a complete binary tree, return the number of the nodes in the tree.
According to Wikipedia, every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level... | [Python/Python3] Count Complete Tree Nodes | 336 | count-complete-tree-nodes | 0.598 | newborncoder | Medium | 3,976 | 222 |
rectangle area | class Solution:
def computeArea(self, ax1: int, ay1: int, ax2: int, ay2: int, bx1: int, by1: int, bx2: int, by2: int) -> int:
coxl=max(ax1,bx1)
coxr=min(ax2,bx2)
coyl=max(ay1,by1)
coyr=min(ay2,by2)
dx=coxr-coxl
dy=coyr-coyl
comm=0
if dx>0 and dy>0:
... | https://leetcode.com/problems/rectangle-area/discuss/2822409/Fastest-python-solution | 9 | Given the coordinates of two rectilinear rectangles in a 2D plane, return the total area covered by the two rectangles.
The first rectangle is defined by its bottom-left corner (ax1, ay1) and its top-right corner (ax2, ay2).
The second rectangle is defined by its bottom-left corner (bx1, by1) and its top-right corner (... | Fastest python solution | 569 | rectangle-area | 0.449 | shubham_1307 | Medium | 4,004 | 223 |
basic calculator | class Solution:
def calculate(self, s: str) -> int:
output, curr, sign, stack = 0, 0, 1, []
for c in s:
if c.isdigit():
curr = (curr * 10) + int(c)
elif c in '+-':
output += curr * sign
curr = 0
if c... | https://leetcode.com/problems/basic-calculator/discuss/2832718/Python-(Faster-than-98)-or-O(N)-stack-solution | 2 | Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.
Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().
Example 1:
Input: s = "1 + 1"
Output: 2
Example 2:
Input: ... | Python (Faster than 98%) | O(N) stack solution | 81 | basic-calculator | 0.423 | KevinJM17 | Hard | 4,054 | 224 |
invert binary tree | class Solution(object):
def invertTree(self, root):
# Base case...
if root == None:
return root
# swapping process...
root.left, root.right = root.right, root.left
# Call the function recursively for the left subtree...
self.invertTree(root.left)
#... | https://leetcode.com/problems/invert-binary-tree/discuss/2463600/Easy-oror-100-oror-Fully-Explained-oror-Java-C%2B%2B-Python-JS-C-Python3-oror-Recursive-and-Iterative | 54 | Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:
Input: root = [2,1,3]
Output: [2,3,1]
Example 3:
Input: root = []
Output: []
Constraints:
The number of nodes in the tree is in the range [0, 100].
-100 <= Node.val <= 1... | Easy || 100% || Fully Explained || Java, C++, Python, JS, C, Python3 || Recursive & Iterative | 3,300 | invert-binary-tree | 0.734 | PratikSen07 | Easy | 4,091 | 226 |
basic calculator ii | class Solution:
def calculate(self, s: str) -> int:
curr_res = 0
res = 0
num = 0
op = "+" # keep the last operator we have seen
# append a "+" sign at the end because we can catch the very last item
for ch in s + "+":
if ch.isdigit():
n... | https://leetcode.com/problems/basic-calculator-ii/discuss/1209116/python-without-any-stack-and-beat-99 | 14 | Given a string s which represents an expression, evaluate this expression and return its value.
The integer division should truncate toward zero.
You may assume that the given expression is always valid. All intermediate results will be in the range of [-231, 231 - 1].
Note: You are not allowed to use any built-in fun... | python - without any stack and beat 99% | 1,300 | basic-calculator-ii | 0.423 | ZAbird | Medium | 4,123 | 227 |
summary ranges | class Solution:
def summaryRanges(self, nums: List[int]) -> List[str]:
start = 0
end = 0
result = []
while start < len(nums) and end<len(nums):
if end+1 < len(nums) and nums[end]+1 == nums[end+1]:
end=end+1
else:
if start == end:
result.append(str(nums[start]))
start = start + 1
... | https://leetcode.com/problems/summary-ranges/discuss/1805476/Python-Simple-Python-Solution-Using-Iterative-Approach-oror-O(n) | 28 | You are given a sorted unique integer array nums.
A range [a,b] is the set of all integers from a to b (inclusive).
Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is i... | [ Python ] ✔✔ Simple Python Solution Using Iterative Approach || O(n) 🔥✌ | 2,300 | summary-ranges | 0.469 | ASHOK_KUMAR_MEGHVANSHI | Easy | 4,149 | 228 |
majority element ii | class Solution:
def majorityElement(self, nums: List[int]) -> List[int]:
return [x for x in set(nums) if nums.count(x) > len(nums)/3] | https://leetcode.com/problems/majority-element-ii/discuss/1483430/Python-95-faster-in-speed | 3 | Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times.
Example 1:
Input: nums = [3,2,3]
Output: [3]
Example 2:
Input: nums = [1]
Output: [1]
Example 3:
Input: nums = [1,2]
Output: [1,2]
Constraints:
1 <= nums.length <= 5 * 104
-109 <= nums[i] <= 109
Follow up: Could you solve the... | Python // 95% faster in speed | 394 | majority-element-ii | 0.442 | fabioo29 | Medium | 4,197 | 229 |
kth smallest element in a bst | class Solution:
import heapq
def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
heap = []
def inorder(r):
if r:
inorder(r.left)
heapq.heappush(heap,-(r.val))
if len(heap) > k:
heapq.heappop(heap... | https://leetcode.com/problems/kth-smallest-element-in-a-bst/discuss/1960632/Inorder-%2B-Heap-In-Python | 2 | Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.
Example 1:
Input: root = [3,1,4,null,2], k = 1
Output: 1
Example 2:
Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3
Constraints:
The number of nodes in the tree is n... | Inorder + Heap In Python | 83 | kth-smallest-element-in-a-bst | 0.695 | gamitejpratapsingh998 | Medium | 4,248 | 230 |
power of two | class Solution:
def isPowerOfTwo(self, n: int) -> bool:
return n>0 and n&(n-1)==0 | https://leetcode.com/problems/power-of-two/discuss/948641/Python-O(1)-Solution | 37 | Given an integer n, return true if it is a power of two. Otherwise, return false.
An integer n is a power of two, if there exists an integer x such that n == 2x.
Example 1:
Input: n = 1
Output: true
Explanation: 20 = 1
Example 2:
Input: n = 16
Output: true
Explanation: 24 = 16
Example 3:
Input: n = 3
Output: false
... | Python O(1) Solution | 2,400 | power-of-two | 0.457 | lokeshsenthilkumar | Easy | 4,273 | 231 |
number of digit one | class Solution:
def countDigitOne(self, n: int) -> int:
#O(logn) mathematical solution
#intervals of new 1s: 0-9, 10-99, 100-999, 1000,9999...
#each interval yields 1,10,100,etc. new '1's respectively
#first and foremost, we want to check how many of each interval repeats
... | https://leetcode.com/problems/number-of-digit-one/discuss/1655517/Python3-O(9)-Straight-Math-Solution | 1 | Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.
Example 1:
Input: n = 13
Output: 6
Example 2:
Input: n = 0
Output: 0
Constraints:
0 <= n <= 109 | Python3 O(9) Straight Math Solution | 341 | number-of-digit-one | 0.342 | terrysu64 | Hard | 4,333 | 233 |
palindrome linked list | class Solution:
def isPalindrome(self, head: Optional[ListNode]) -> bool:
def reverse(node):
prev = None
while node:
next_node = node.next
node.next = prev
prev, node = node, next_node
return prev
... | https://leetcode.com/problems/palindrome-linked-list/discuss/2466200/Python-O(N)O(1) | 7 | Given the head of a singly linked list, return true if it is a
palindrome
or false otherwise.
Example 1:
Input: head = [1,2,2,1]
Output: true
Example 2:
Input: head = [1,2]
Output: false
Constraints:
The number of nodes in the list is in the range [1, 105].
0 <= Node.val <= 9
Follow up: Could you do it in O(n) ti... | Python, O(N)/O(1) | 839 | palindrome-linked-list | 0.496 | blue_sky5 | Easy | 4,344 | 234 |
lowest common ancestor of a binary search tree | class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
while True:
if root.val > p.val and root.val > q.val:
root = root.left
elif root.val < p.val and root.val < q.val:
root = root.right
... | https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/discuss/1394823/Explained-Easy-Iterative-Python-Solution | 55 | Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descen... | Explained Easy Iterative Python Solution | 2,000 | lowest-common-ancestor-of-a-binary-search-tree | 0.604 | sevdariklejdi | Medium | 4,377 | 235 |
lowest common ancestor of a binary tree | class Solution:
def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
if root==None or root.val==p.val or root.val==q.val:
return root
left=self.lowestCommonAncestor(root.left,p,q)
right=self.lowestCommonAncestor(root.right,p,q)
if ... | https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/discuss/2539410/python3-simple-Solution | 4 | Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”... | python3 simple Solution | 251 | lowest-common-ancestor-of-a-binary-tree | 0.581 | pranjalmishra334 | Medium | 4,399 | 236 |
delete node in a linked list | class Solution:
def deleteNode(self, node):
nextNode = node.next
node.val = nextNode.val
node.next = nextNode.next | https://leetcode.com/problems/delete-node-in-a-linked-list/discuss/1454184/95.96-faster-and-Simpler-solution-with-Explanation. | 6 | There is a singly-linked list head and we want to delete a node node in it.
You are given the node to be deleted node. You will not be given access to the first node of head.
All the values of the linked list are unique, and it is guaranteed that the given node node is not the last node in the linked list.
Delete the g... | 95.96% faster and Simpler solution with Explanation. | 870 | delete-node-in-a-linked-list | 0.753 | AmrinderKaur1 | Medium | 4,427 | 237 |
product of array except self | class Solution:
def productExceptSelf(self, nums: List[int]) -> List[int]:
leftProducts = [0]*len(nums) # initialize left array
rightProducts = [0]*len(nums) # initialize right array
leftProducts[0] = 1 # the left most is 1
rightProducts[-1] = 1 # the right most is 1
... | https://leetcode.com/problems/product-of-array-except-self/discuss/744951/Product-of-array-except-self-Python3-Solution-with-a-Detailed-Explanation | 12 | Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i].
The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer.
You must write an algorithm that runs in O(n) time and without using the division operation.
... | Product of array except self - Python3 Solution with a Detailed Explanation | 1,200 | product-of-array-except-self | 0.648 | peyman_np | Medium | 4,452 | 238 |
sliding window maximum | class Solution:
def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]:
res = []
window = collections.deque()
for i, num in enumerate(nums):
while window and num >= nums[window[-1]]:
window.pop()
window.append(i)
if i... | https://leetcode.com/problems/sliding-window-maximum/discuss/1624633/Python-3-O(n) | 3 | You are given an array of integers nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.
Return the max sliding window.
Example 1:
Input: nums = [1,3,-1,-3,5,3... | Python 3 O(n) | 772 | sliding-window-maximum | 0.466 | dereky4 | Hard | 4,510 | 239 |
search a 2d matrix ii | class Solution:
def searchMatrix(self, mat: List[List[int]], target: int) -> bool:
m=len(mat)
n=len(mat[0])
i=m-1
j=0
while i>=0 and j<n:
if mat[i][j]==target:
return True
elif mat[i][j]<target:
... | https://leetcode.com/problems/search-a-2d-matrix-ii/discuss/2324351/PYTHON-oror-EXPLAINED-oror | 48 | Write an efficient algorithm that searches for a value target in an m x n integer matrix matrix. This matrix has the following properties:
Integers in each row are sorted in ascending from left to right.
Integers in each column are sorted in ascending from top to bottom.
Example 1:
Input: matrix = [[1,4,7,11,15],[2,5... | ✔️ PYTHON || EXPLAINED || ; ] | 1,700 | search-a-2d-matrix-ii | 0.507 | karan_8082 | Medium | 4,542 | 240 |
different ways to add parentheses | class Solution(object):
def diffWaysToCompute(self, s, memo=dict()):
if s in memo:
return memo[s]
if s.isdigit(): # base case
return [int(s)]
calculate = {'*': lambda x, y: x * y,
'+': lambda x, y: x + y,
'-': lambda x, y: x -... | https://leetcode.com/problems/different-ways-to-add-parentheses/discuss/2392799/Python3-Divide-and-Conquer%3A-Recursion-%2B-Memoization | 4 | Given a string expression of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. You may return the answer in any order.
The test cases are generated such that the output values fit in a 32-bit integer and the number of different results does... | [Python3] Divide and Conquer: Recursion + Memoization | 156 | different-ways-to-add-parentheses | 0.633 | mhpd | Medium | 4,589 | 241 |
valid anagram | class Solution:
def isAnagram(self, s: str, t: str) -> bool:
tracker = collections.defaultdict(int)
for x in s: tracker[x] += 1
for x in t: tracker[x] -= 1
return all(x == 0 for x in tracker.values()) | https://leetcode.com/problems/valid-anagram/discuss/433680/Python-3-O(n)-Faster-than-98.39-Memory-usage-less-than-100 | 69 | Given two strings s and t, return true if t is an anagram of s, and false otherwise.
An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.
Example 1:
Input: s = "anagram", t = "nagaram"
Output: true
Example 2:
Input: s = ... | Python 3 - O(n) - Faster than 98.39%, Memory usage less than 100% | 18,500 | valid-anagram | 0.628 | mmbhatk | Easy | 4,618 | 242 |
binary tree paths | class Solution:
def binaryTreePaths(self, R: TreeNode) -> List[str]:
A, P = [], []
def dfs(N):
if N == None: return
P.append(N.val)
if (N.left,N.right) == (None,None): A.append('->'.join(map(str,P)))
else: dfs(N.left), dfs(N.right)
P.pop()
... | https://leetcode.com/problems/binary-tree-paths/discuss/484118/Python-3-(beats-~100)-(nine-lines)-(DFS) | 7 | Given the root of a binary tree, return all root-to-leaf paths in any order.
A leaf is a node with no children.
Example 1:
Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]
Example 2:
Input: root = [1]
Output: ["1"]
Constraints:
The number of nodes in the tree is in the range [1, 100].
-100 <= Node.val <= 100 | Python 3 (beats ~100%) (nine lines) (DFS) | 1,100 | binary-tree-paths | 0.607 | junaidmansuri | Easy | 4,672 | 257 |
add digits | class Solution(object):
def addDigits(self, num):
while num > 9:
num = num % 10 + num // 10
return num | https://leetcode.com/problems/add-digits/discuss/2368005/Very-Easy-100-(Fully-Explained)(Java-C%2B%2B-Python-JS-C-Python3) | 13 | Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.
Example 1:
Input: num = 38
Output: 2
Explanation: The process is
38 --> 3 + 8 --> 11
11 --> 1 + 1 --> 2
Since 2 has only one digit, return it.
Example 2:
Input: num = 0
Output: 0
Constraints:
0 <= num <= 231 - 1... | Very Easy 100% (Fully Explained)(Java, C++, Python, JS, C, Python3) | 484 | add-digits | 0.635 | PratikSen07 | Easy | 4,696 | 258 |
single number iii | class Solution:
def singleNumber(self, nums: List[int]) -> List[int]:
dc=defaultdict(lambda:0)
for a in(nums):
dc[a]+=1
ans=[]
for a in dc:
if(dc[a]==1):
ans.append(a)
return ans | https://leetcode.com/problems/single-number-iii/discuss/2828366/brute-force-with-dictnory | 5 | Given an integer array nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once. You can return the answer in any order.
You must write an algorithm that runs in linear runtime complexity and uses only constant extra space.
Examp... | brute force with dictnory | 30 | single-number-iii | 0.675 | droj | Medium | 4,754 | 260 |
ugly number | class Solution:
def isUgly(self, num: int) -> bool:
if num == 0: return False
while num % 5 == 0: num /= 5
while num % 3 == 0: num /= 3
while num % 2 == 0: num /= 2
return num == 1
- Junaid Mansuri | https://leetcode.com/problems/ugly-number/discuss/336227/Solution-in-Python-3-(beats-~99)-(five-lines) | 18 | An ugly number is a positive integer whose prime factors are limited to 2, 3, and 5.
Given an integer n, return true if n is an ugly number.
Example 1:
Input: n = 6
Output: true
Explanation: 6 = 2 × 3
Example 2:
Input: n = 1
Output: true
Explanation: 1 has no prime factors, therefore all of its prime factors are limi... | Solution in Python 3 (beats ~99%) (five lines) | 2,600 | ugly-number | 0.426 | junaidmansuri | Easy | 4,785 | 263 |
ugly number ii | class Solution:
def nthUglyNumber(self, n: int) -> int:
k = [0] * n
t1 = t2 = t3 = 0
k[0] = 1
for i in range(1,n):
k[i] = min(k[t1]*2,k[t2]*3,k[t3]*5)
if(k[i] == k[t1]*2): t1 += 1
if(k[i] == k[t2]*3): t2 += 1
if(k[i] == k[t3]*5): t3 += ... | https://leetcode.com/problems/ugly-number-ii/discuss/556314/Python-Simple-DP-9-Lines | 23 | An ugly number is a positive integer whose prime factors are limited to 2, 3, and 5.
Given an integer n, return the nth ugly number.
Example 1:
Input: n = 10
Output: 12
Explanation: [1, 2, 3, 4, 5, 6, 8, 9, 10, 12] is the sequence of the first 10 ugly numbers.
Example 2:
Input: n = 1
Output: 1
Explanation: 1 has no p... | [Python] Simple DP 9 Lines | 1,100 | ugly-number-ii | 0.462 | mazz272727 | Medium | 4,839 | 264 |
missing number | class Solution:
def missingNumber(self, nums: List[int]) -> int:
return (len(nums) * (len(nums) + 1))//2 - sum(nums) | https://leetcode.com/problems/missing-number/discuss/2081185/Python-Easy-One-liners-with-Explanation | 23 | Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.
Example 1:
Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does ... | ✅ Python Easy One liners with Explanation | 2,000 | missing-number | 0.617 | constantine786 | Easy | 4,860 | 268 |
integer to english words | class Solution:
def numberToWords(self, num: int) -> str:
mp = {1: "One", 11: "Eleven", 10: "Ten",
2: "Two", 12: "Twelve", 20: "Twenty",
3: "Three", 13: "Thirteen", 30: "Thirty",
4: "Four", 14: "Fourteen", 40: "Forty",
5: "Five", 15: "... | https://leetcode.com/problems/integer-to-english-words/discuss/1990823/JavaC%2B%2BPythonJavaScriptKotlinSwiftO(n)timeBEATS-99.97-MEMORYSPEED-0ms-APRIL-2022 | 8 | Convert a non-negative integer num to its English words representation.
Example 1:
Input: num = 123
Output: "One Hundred Twenty Three"
Example 2:
Input: num = 12345
Output: "Twelve Thousand Three Hundred Forty Five"
Example 3:
Input: num = 1234567
Output: "One Million Two Hundred Thirty Four Thousand Five Hundred Six... | [Java/C++/Python/JavaScript/Kotlin/Swift]O(n)time/BEATS 99.97% MEMORY/SPEED 0ms APRIL 2022 | 677 | integer-to-english-words | 0.299 | cucerdariancatalin | Hard | 4,927 | 273 |
h index | class Solution:
def hIndex(self, citations: List[int]) -> int:
tmp = [0] * (len(citations) + 1)
for i in range(len(citations)):
if citations[i] > len(citations):
tmp[len(citations)] += 1
else:
tmp[citations[i]] += 1
sum_ = 0
... | https://leetcode.com/problems/h-index/discuss/785586/Python3-O(n)-without-sorting! | 5 | Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper, return the researcher's h-index.
According to the definition of h-index on Wikipedia: The h-index is defined as the maximum value of h such that the given researcher has published at least h pap... | Python3 O(n) without sorting! | 489 | h-index | 0.382 | DebbieAlter | Medium | 4,949 | 274 |
h index ii | class Solution:
def hIndex(self, A):
n = len(A)
l, r = 0, n - 1
while l < r:
m = (l + r + 1) // 2
if A[m] > n - m: r = m - 1
else: l = m
return n - l - (A[l] < n - l) | https://leetcode.com/problems/h-index-ii/discuss/2729382/Python3-Solution-or-Binary-Search-or-O(logn) | 2 | Given an array of integers citations where citations[i] is the number of citations a researcher received for their ith paper and citations is sorted in ascending order, return the researcher's h-index.
According to the definition of h-index on Wikipedia: The h-index is defined as the maximum value of h such that the gi... | ✔ Python3 Solution | Binary Search | O(logn) | 109 | h-index-ii | 0.374 | satyam2001 | Medium | 4,965 | 275 |
first bad version | class Solution:
def firstBadVersion(self, n: int) -> int:
left = 1
right = n
result = 1
while left<=right:
mid = (left+right)//2
if isBadVersion(mid) == False:
left = mid+1
else:
right = mid-1
... | https://leetcode.com/problems/first-bad-version/discuss/2700688/Simple-Python-Solution-Using-Binary-Search | 25 | You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad.
Suppose you have n versions [1, 2, ..., n] and you... | ✔️ Simple Python Solution Using Binary Search 🔥 | 3,400 | first-bad-version | 0.43 | pniraj657 | Easy | 4,979 | 278 |
perfect squares | class Solution:
def isSquare(self, n: int) -> bool:
sq = int(math.sqrt(n))
return sq*sq == n
def numSquares(self, n: int) -> int:
# Lagrange's four-square theorem
if self.isSquare(n):
return 1
while (n & 3) == 0:
n >>= 2
if (n ... | https://leetcode.com/problems/perfect-squares/discuss/376795/100-O(log-n)-Python3-Solution-Lagrange's-four-square-theorem | 12 | Given an integer n, return the least number of perfect square numbers that sum to n.
A perfect square is an integer that is the square of an integer; in other words, it is the product of some integer with itself. For example, 1, 4, 9, and 16 are perfect squares while 3 and 11 are not.
Example 1:
Input: n = 12
Output:... | 100% O(log n) Python3 Solution - Lagrange’s four-square theorem | 1,200 | perfect-squares | 0.526 | TCarmic | Medium | 4,993 | 279 |
expression add operators | class Solution:
def addOperators(self, num: str, target: int) -> List[str]:
exprs = []
def recurse(idx, value, delta, exp):
# base case here
if idx == len(num):
if value == target:
exprs.append("".join(exp))
... | https://leetcode.com/problems/expression-add-operators/discuss/1031229/Python-Simple-heavily-commented-and-accepted-Recursive-Solution | 4 | Given a string num that contains only digits and an integer target, return all possibilities to insert the binary operators '+', '-', and/or '*' between the digits of num so that the resultant expression evaluates to the target value.
Note that operands in the returned expressions should not contain leading zeros.
Ex... | [Python] Simple, heavily commented and accepted Recursive Solution | 354 | expression-add-operators | 0.392 | gokivego | Hard | 5,046 | 282 |
move zeroes | class Solution(object):
def moveZeroes(self, nums):
i=0
n = len(nums)
while i <n:
if nums[i]==0:
nums.pop(i)
nums.append(0)
n-=1
else:
i+=1 | https://leetcode.com/problems/move-zeroes/discuss/404010/Python-easy-solution | 16 | Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.
Note that you must do this in-place without making a copy of the array.
Example 1:
Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]
Example 2:
Input: nums = [0]
Output: [0]
Constraints:
1 <= nums... | Python easy solution | 2,600 | move-zeroes | 0.614 | saffi | Easy | 5,056 | 283 |
find the duplicate number | class Solution:
def findDuplicate(self, nums: List[int]) -> int:
slow = fast = ans = 0
while True:
slow = nums[slow]
fast = nums[nums[fast]]
if slow == fast:
break
while ans != slow:
ans = nums[ans]
slow = nums[slow]... | https://leetcode.com/problems/find-the-duplicate-number/discuss/650942/Proof-of-Floyd's-cycle-detection-algorithm-Find-the-Duplicate-Number | 20 | Given an array of integers nums containing n + 1 integers where each integer is in the range [1, n] inclusive.
There is only one repeated number in nums, return this repeated number.
You must solve the problem without modifying the array nums and uses only constant extra space.
Example 1:
Input: nums = [1,3,4,2,2]
Ou... | Proof of Floyd's cycle detection algorithm - Find the Duplicate Number | 2,400 | find-the-duplicate-number | 0.591 | r0bertz | Medium | 5,101 | 287 |
game of life | class Solution:
def gameOfLife(self, board: List[List[int]]) -> None:
"""
Do not return anything, modify board in-place instead.
"""
def is_neighbor(board, i, j):
return (0 <= i < len(board)) and (0 <= j < len(board[0])) and board[i][j] % 10 == 1
... | https://leetcode.com/problems/game-of-life/discuss/468108/Use-the-tens-digit-as-a-counter-Python-O(1)-Space-O(mn)-Time | 4 | According to Wikipedia's article: "The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."
The board is made up of an m x n grid of cells, where each cell has an initial state: live (represented by a 1) or dead (represented by a 0). Each cel... | Use the tens digit as a counter, Python O(1) Space, O(mn) Time | 104 | game-of-life | 0.668 | Moooooon | Medium | 5,156 | 289 |
word pattern | class Solution:
def wordPattern(self, pattern: str, s: str) -> bool:
li = s.split(' ')
di = {}
if len(li) != len(pattern):
return False
for i, val in enumerate(pattern):
if val in di and di[val] != li[i]:
return False
elif ... | https://leetcode.com/problems/word-pattern/discuss/1696590/Simple-Python-Solution-oror-Faster-than-99.34 | 4 | Given a pattern and a string s, find if s follows the same pattern.
Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in s.
Example 1:
Input: pattern = "abba", s = "dog cat cat dog"
Output: true
Example 2:
Input: pattern = "abba", s = "dog cat cat fish"
Ou... | Simple Python Solution || Faster than 99.34% | 509 | word-pattern | 0.404 | KiranUpase | Easy | 5,202 | 290 |
nim game | class Solution:
def canWinNim(self, n: int) -> bool:
if n <= 3:
return True
new_size = n + 1
memo = [False] * (new_size)
for i in range(4):
memo[i] = True
for i in range(4,new_size):
for j in range(1,4):
... | https://leetcode.com/problems/nim-game/discuss/1141120/Bottom-up-DP-python | 9 | You are playing the following Nim Game with your friend:
Initially, there is a heap of stones on the table.
You and your friend will alternate taking turns, and you go first.
On each turn, the person whose turn it is will remove 1 to 3 stones from the heap.
The one who removes the last stone is the winner.
Given n, the... | Bottom-up DP python | 725 | nim-game | 0.559 | lywc | Easy | 5,249 | 292 |
bulls and cows | class Solution:
def getHint(self, secret: str, guess: str) -> str:
# The main idea is to understand that cow cases contain the bull cases
# This loop will take care of "bull" cases
bull=0
for i in range(len(secret)):
bull += int(secret[i] == guess[i])
# This lo... | https://leetcode.com/problems/bulls-and-cows/discuss/563661/Fast-and-easy-to-understand-Python-solution-O(n) | 14 | You are playing the Bulls and Cows game with your friend.
You write down a secret number and ask your friend to guess what the number is. When your friend makes a guess, you provide a hint with the following info:
The number of "bulls", which are digits in the guess that are in the correct position.
The number of "cows... | Fast and easy to understand Python solution O(n) | 983 | bulls-and-cows | 0.487 | mista2311 | Medium | 5,269 | 299 |
longest increasing subsequence | class Solution: # Suppose, for example:
# nums = [1,8,4,5,3,7],
# for which the longest strictly increasing subsequence is arr = [1,4,5,7],
# giving len(arr) = 4 as the answer
#
# Here's the plan:
... | https://leetcode.com/problems/longest-increasing-subsequence/discuss/2395570/Python3-oror-7-lines-binSearch-cheating-wexplanation-oror-TM%3A-9482 | 45 | Given an integer array nums, return the length of the longest strictly increasing
subsequence
.
Example 1:
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = ... | Python3 || 7 lines, binSearch, cheating, w/explanation || T/M: 94%/82% | 4,100 | longest-increasing-subsequence | 0.516 | warrenruud | Medium | 5,303 | 300 |
remove invalid parentheses | class Solution:
def removeInvalidParentheses(self, s: str) -> List[str]:
def valid(s):
l,r=0,0
for c in s:
if c=='(':
l+=1
elif c==')':
if l<=0:
r+=1
else:
... | https://leetcode.com/problems/remove-invalid-parentheses/discuss/1555755/Easy-to-understand-Python-solution | 3 | Given a string s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.
Return a list of unique strings that are valid with the minimum number of removals. You may return the answer in any order.
Example 1:
Input: s = "()())()"
Output: ["(())()","()()()... | Easy to understand Python 🐍 solution | 375 | remove-invalid-parentheses | 0.471 | InjySarhan | Hard | 5,356 | 301 |
additive number | class Solution:
def isAdditiveNumber(self, s: str) -> bool:
n = len(s)
for i in range(1, n): # Choose the length of first number
# If length of 1st number is > 1 and starts with 0 -- skip
if i != 1 and s[0] == '0':
continue
for j in range(1, n): # ... | https://leetcode.com/problems/additive-number/discuss/2764371/Python-solution-with-complete-explanation-and-code | 0 | An additive number is a string whose digits can form an additive sequence.
A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.
Given a string containing only digits, return true if it is an additi... | Python solution with complete explanation and code | 11 | additive-number | 0.309 | smit5300 | Medium | 5,364 | 306 |
best time to buy and sell stock with cooldown | class Solution:
def maxProfit(self, prices: List[int]) -> int:
n = len(prices)
dp = [[0 for i in range(2)] for i in range(n+2)]
dp[n][0] = dp[n][1] = 0
ind = n-1
while(ind>=0):
for buy in range(2):
if(buy):
... | https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/discuss/2132119/Python-or-Easy-DP-or | 4 | You are given an array prices where prices[i] is the price of a given stock on the ith day.
Find the maximum profit you can achieve. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions:
After you sell your stock, you cannot buy... | Python | Easy DP | | 166 | best-time-to-buy-and-sell-stock-with-cooldown | 0.546 | LittleMonster23 | Medium | 5,369 | 309 |
minimum height trees | class Solution:
def findMinHeightTrees(self, n: int, edges: List[List[int]]) -> List[int]:
if n == 1:
return [0]
graph = {i:[] for i in range(n)}
for u, v in edges:
graph[u].append(v)
graph[v].append(u)
leaves = []
for node in grap... | https://leetcode.com/problems/minimum-height-trees/discuss/1753794/Python-easy-to-read-and-understand-or-reverse-topological-sort | 7 | A tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.
Given a tree of n nodes labelled from 0 to n - 1, and an array of n - 1 edges where edges[i] = [ai, bi] indicates that there is an undirected edge between the t... | Python easy to read and understand | reverse topological sort | 336 | minimum-height-trees | 0.385 | sanial2001 | Medium | 5,399 | 310 |
burst balloons | class Solution(object):
def maxCoins(self, nums):
n=len(nums)
nums.insert(n,1)
nums.insert(0,1)
self.dp={}
return self.dfs(1,nums,n)
def dfs(self,strt,nums,end):
ans=0
if strt>end:
return 0
if (strt,end) in self.dp:
return s... | https://leetcode.com/problems/burst-balloons/discuss/1477014/Python3-or-Top-down-Approach | 2 | You are given n balloons, indexed from 0 to n - 1. Each balloon is painted with a number on it represented by an array nums. You are asked to burst all the balloons.
If you burst the ith balloon, you will get nums[i - 1] * nums[i] * nums[i + 1] coins. If i - 1 or i + 1 goes out of bounds of the array, then treat it as ... | [Python3] | Top-down Approach | 269 | burst-balloons | 0.568 | swapnilsingh421 | Hard | 5,412 | 312 |
super ugly number | class Solution:
def nthSuperUglyNumber(self, n: int, primes: List[int]) -> int:
hp=[1]
dc={1}
i=1
while(n):
mn=heapq.heappop(hp)
if(n==1):
return mn
for p in primes:
newno=mn*p
if(newno in dc):
... | https://leetcode.com/problems/super-ugly-number/discuss/2828266/heapq-did-the-job | 5 | A super ugly number is a positive integer whose prime factors are in the array primes.
Given an integer n and an array of integers primes, return the nth super ugly number.
The nth super ugly number is guaranteed to fit in a 32-bit signed integer.
Example 1:
Input: n = 12, primes = [2,7,13,19]
Output: 32
Explanation:... | heapq did the job | 40 | super-ugly-number | 0.458 | droj | Medium | 5,421 | 313 |
count of smaller numbers after self | class Solution: # Here's the plan:
# 1) Make arr, a sorted copy of the list nums.
# 2) iterate through nums. For each element num in nums:
# 2a) use a binary search to determine the count of elements
# in the arr that ... | https://leetcode.com/problems/count-of-smaller-numbers-after-self/discuss/2320000/Python3.-oror-binSearch-6-lines-w-explanation-oror-TM%3A-9784 | 27 | Given an integer array nums, return an integer array counts where counts[i] is the number of smaller elements to the right of nums[i].
Example 1:
Input: nums = [5,2,6,1]
Output: [2,1,1,0]
Explanation:
To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To t... | Python3. || binSearch 6 lines, w/ explanation || T/M: 97%/84% | 3,000 | count-of-smaller-numbers-after-self | 0.428 | warrenruud | Hard | 5,429 | 315 |
remove duplicate letters | class Solution:
def removeDuplicateLetters(self, s: str) -> str:
stack = []
for idx, character in enumerate(s):
if not stack:
stack.append(character)
elif character in stack:
continue
else:
while stack and (... | https://leetcode.com/problems/remove-duplicate-letters/discuss/1687144/Python-3-Simple-solution-using-a-stack-and-greedy-approach-(32ms-14.4MB) | 5 | Given a string s, remove duplicate letters so that every letter appears once and only once. You must make sure your result is
the smallest in lexicographical order
among all possible results.
Example 1:
Input: s = "bcabc"
Output: "abc"
Example 2:
Input: s = "cbacdcbc"
Output: "acdb"
Constraints:
1 <= s.length <= 10... | [Python 3] Simple solution using a stack and greedy approach (32ms, 14.4MB) | 498 | remove-duplicate-letters | 0.446 | seankala | Medium | 5,443 | 316 |
maximum product of word lengths | class Solution:
def maxProduct(self, words: List[str]) -> int:
n=len(words)
char_set = [set(words[i]) for i in range(n)] # precompute hashset for each word
max_val = 0
for i in range(n):
for j in ra... | https://leetcode.com/problems/maximum-product-of-word-lengths/discuss/2085316/Python-or-or-Easy-3-Approaches-explained | 57 | Given a string array words, return the maximum value of length(word[i]) * length(word[j]) where the two words do not share common letters. If no such two words exist, return 0.
Example 1:
Input: words = ["abcw","baz","foo","bar","xtfn","abcdef"]
Output: 16
Explanation: The two words can be "abcw", "xtfn".
Example 2:
... | ✅ Python | | Easy 3 Approaches explained | 3,700 | maximum-product-of-word-lengths | 0.601 | constantine786 | Medium | 5,463 | 318 |
bulb switcher | class Solution:
def bulbSwitch(self, n: int) -> int:
# Only those bulds with perferct square number index will keep "ON" at last.
return int(n**0.5) | https://leetcode.com/problems/bulb-switcher/discuss/535399/PythonJSJavaC%2B%2B-sol-by-perfect-square.-w-Visualization | 20 | There are n bulbs that are initially off. You first turn on all the bulbs, then you turn off every second bulb.
On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb.
Return the num... | Python/JS/Java/C++ sol by perfect square. [w/ Visualization] | 1,200 | bulb-switcher | 0.481 | brianchiang_tw | Medium | 5,507 | 319 |
create maximum number | class Solution:
def maxNumber(self, nums1: List[int], nums2: List[int], k: int) -> List[int]:
def maximum_num_each_list(nums: List[int], k_i: int) -> List[int]:
# monotonically decreasing stack
s = []
m = len(nums) - k_i
for n in nums:
while s ... | https://leetcode.com/problems/create-maximum-number/discuss/2167532/O(k(m%2Bn)).-Python-Solution-with-monotonically-decreasing-stack.-Commented-for-clarity. | 2 | You are given two integer arrays nums1 and nums2 of lengths m and n respectively. nums1 and nums2 represent the digits of two numbers. You are also given an integer k.
Create the maximum number of length k <= m + n from digits of the two numbers. The relative order of the digits from the same array must be preserved.
R... | O(k(m+n)). Python Solution with monotonically decreasing stack. Commented for clarity. | 443 | create-maximum-number | 0.288 | saqibmubarak | Hard | 5,514 | 321 |
coin change | class Solution:
def coinChange(self, coins: List[int], amount: int) -> int:
dp=[math.inf] * (amount+1)
dp[0]=0
for coin in coins:
for i in range(coin, amount+1):
if i-coin>=0:
dp[i]=min(dp[i], dp[i-coin]+1)
ret... | https://leetcode.com/problems/coin-change/discuss/2058537/Python-Easy-2-DP-approaches | 32 | You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.
Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.
You may assume that... | Python Easy 2 DP approaches | 3,800 | coin-change | 0.416 | constantine786 | Medium | 5,519 | 322 |
wiggle sort ii | class Solution:
def wiggleSort(self, nums: List[int]) -> None:
sortedList = sorted(nums)
n = len(nums)
if n%2==0:
small = sortedList[:((n//2))][::-1]
large = (sortedList[(n//2):])[::-1]
for i in range(1,n,2):
nums[i] = large[i//2]
... | https://leetcode.com/problems/wiggle-sort-ii/discuss/1322709/Definitely-not-O(n)-but-did-it-iteratively-in-O(nlog(N))-time | 1 | Given an integer array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3]....
You may assume the input array always has a valid answer.
Example 1:
Input: nums = [1,5,1,1,6,4]
Output: [1,6,1,5,1,4]
Explanation: [1,4,1,5,1,6] is also accepted.
Example 2:
Input: nums = [1,3,2,2,3,1]
Output: [2,3,1,3,1,2]
... | Definitely not O(n) but did it iteratively in O(nlog(N)) time | 180 | wiggle-sort-ii | 0.33 | prajwalPonnana004 | Medium | 5,579 | 324 |
power of three | class Solution:
def isPowerOfThree(self, n: int) -> bool:
if n == 1:
return True
if n == 0:
return False
else:
return n % 3 == 0 and self.isPowerOfThree(n // 3) | https://leetcode.com/problems/power-of-three/discuss/1179790/Simple-Python-Recursive-Solution-with-Explanation | 8 | Given an integer n, return true if it is a power of three. Otherwise, return false.
An integer n is a power of three, if there exists an integer x such that n == 3x.
Example 1:
Input: n = 27
Output: true
Explanation: 27 = 33
Example 2:
Input: n = 0
Output: false
Explanation: There is no x where 3x = 0.
Example 3:
Inp... | Simple Python Recursive Solution with Explanation | 356 | power-of-three | 0.453 | stevenbooke | Easy | 5,582 | 326 |
count of range sum | class Solution:
def countRangeSum(self, nums: List[int], lower: int, upper: int) -> int:
prefix = [0]
for x in nums: prefix.append(prefix[-1] + x)
def fn(lo, hi):
"""Return count of range sum between prefix[lo:hi]."""
if lo+1 >= hi: return 0
mid... | https://leetcode.com/problems/count-of-range-sum/discuss/1195369/Python3-4-solutions | 8 | Given an integer array nums and two integers lower and upper, return the number of range sums that lie in [lower, upper] inclusive.
Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j inclusive, where i <= j.
Example 1:
Input: nums = [-2,5,-1], lower = -2, upper = 2
Output: 3
Expla... | [Python3] 4 solutions | 459 | count-of-range-sum | 0.361 | ye15 | Hard | 5,643 | 327 |
odd even linked list | class Solution:
def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]:
if(head is None or head.next is None):
return head
# assign odd = head(starting node of ODD)
# assign even = head.next(starting node of EVEN)
odd , even = head , head.next
... | https://leetcode.com/problems/odd-even-linked-list/discuss/2458296/easy-approach-using-two-pointer-in-python-With-Comments-TC-%3A-O(N)-SC-%3A-O(1) | 4 | Given the head of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return the reordered list.
The first node is considered odd, and the second node is even, and so on.
Note that the relative order inside both the even and odd groups should remain as it was... | easy approach using two pointer in python With Comments [TC : O(N) SC : O(1)] | 156 | odd-even-linked-list | 0.603 | rajitkumarchauhan99 | Medium | 5,648 | 328 |
longest increasing path in a matrix | class Solution:
def longestIncreasingPath(self, grid: List[List[int]]) -> int:
m,n=len(grid),len(grid[0])
directions = [0, 1, 0, -1, 0] # four directions
@lru_cache(maxsize=None) # using python cache lib for memoization
def dfs(r,c):
ans=1
#... | https://leetcode.com/problems/longest-increasing-path-in-a-matrix/discuss/2052380/Python-DFS-with-Memoization-Beats-~90 | 4 | Given an m x n integers matrix, return the length of the longest increasing path in matrix.
From each cell, you can either move in four directions: left, right, up, or down. You may not move diagonally or move outside the boundary (i.e., wrap-around is not allowed).
Example 1:
Input: matrix = [[9,9,4],[6,6,8],[2,1,1]... | Python DFS with Memoization Beats ~90% | 580 | longest-increasing-path-in-a-matrix | 0.522 | constantine786 | Hard | 5,669 | 329 |
patching array | class Solution:
def minPatches(self, nums: List[int], n: int) -> int:
ans, total = 0, 0
num_idx = 0
while total < n:
if num_idx < len(nums):
if total < nums[num_idx] - 1:
total = total * 2 + 1
ans += 1
else:
total += nums[num_idx]
num_idx += 1
else:
total = total * 2 + 1
... | https://leetcode.com/problems/patching-array/discuss/1432390/Python-3-easy-solution | 1 | Given a sorted integer array nums and an integer n, add/patch elements to the array such that any number in the range [1, n] inclusive can be formed by the sum of some elements in the array.
Return the minimum number of patches required.
Example 1:
Input: nums = [1,3], n = 6
Output: 1
Explanation:
Combinations of num... | Python 3 easy solution | 231 | patching-array | 0.4 | Andy_Feng97 | Hard | 5,714 | 330 |
verify preorder serialization of a binary tree | class Solution:
def isValidSerialization(self, preorder: str) -> bool:
stack = []
items = preorder.split(",")
for i, val in enumerate(items):
if i>0 and not stack:
return False
if stack:
stack[-1][1] -= 1
if stack[-1][1]... | https://leetcode.com/problems/verify-preorder-serialization-of-a-binary-tree/discuss/1459342/Python3-or-O(n)-Time-and-O(n)-space | 3 | One way to serialize a binary tree is to use preorder traversal. When we encounter a non-null node, we record the node's value. If it is a null node, we record using a sentinel value such as '#'.
For example, the above binary tree can be serialized to the string "9,3,4,#,#,1,#,#,2,#,6,#,#", where '#' represents a null ... | Python3 | O(n) Time and O(n) space | 73 | verify-preorder-serialization-of-a-binary-tree | 0.443 | Sanjaychandak95 | Medium | 5,716 | 331 |
reconstruct itinerary | class Solution:
def __init__(self):
self.path = []
def findItinerary(self, tickets: List[List[str]]) -> List[str]:
flights = {}
# as graph is directed
# => no bi-directional paths
for t1, t2 in tickets:
if t1 not in flights:
fligh... | https://leetcode.com/problems/reconstruct-itinerary/discuss/1475876/Python-LC-but-better-explained | 1 | You are given a list of airline tickets where tickets[i] = [fromi, toi] represent the departure and the arrival airports of one flight. Reconstruct the itinerary in order and return it.
All of the tickets belong to a man who departs from "JFK", thus, the itinerary must begin with "JFK". If there are multiple valid itin... | Python LC, but better explained | 213 | reconstruct-itinerary | 0.41 | SleeplessChallenger | Hard | 5,724 | 332 |
increasing triplet subsequence | class Solution:
def increasingTriplet(self, nums: List[int]) -> bool:
n = len(nums)
maxRight = [0] * n # maxRight[i] is the maximum element among nums[i+1...n-1]
maxRight[-1] = nums[-1]
for i in range(n-2, -1, -1):
maxRight[i] = max(maxRight[i+1], nums[i+1])
... | https://leetcode.com/problems/increasing-triplet-subsequence/discuss/270884/Python-2-solutions%3A-Right-So-Far-One-pass-O(1)-Space-Clean-and-Concise | 94 | Given an integer array nums, return true if there exists a triple of indices (i, j, k) such that i < j < k and nums[i] < nums[j] < nums[k]. If no such indices exists, return false.
Example 1:
Input: nums = [1,2,3,4,5]
Output: true
Explanation: Any triplet where i < j < k is valid.
Example 2:
Input: nums = [5,4,3,2,1]... | [Python] 2 solutions: Right So Far, One pass - O(1) Space - Clean & Concise | 2,000 | increasing-triplet-subsequence | 0.427 | hiepit | Medium | 5,735 | 334 |
self crossing | class Solution:
def isSelfCrossing(self, x: List[int]) -> bool:
def intersect(p1, p2, p3, p4):
v1 = p2 - p1
if v1.real == 0:
return p1.imag <= p3.imag <= p2.imag and p3.real <= p1.real <= p4.real
return p3.imag <= p1.imag <= p4.imag and p1.real <= p3.real ... | https://leetcode.com/problems/self-crossing/discuss/710582/Python3-complex-number-solution-Self-Crossing | 0 | You are given an array of integers distance.
You start at the point (0, 0) on an X-Y plane, and you move distance[0] meters to the north, then distance[1] meters to the west, distance[2] meters to the south, distance[3] meters to the east, and so on. In other words, after each move, your direction changes counter-clock... | Python3 complex number solution - Self Crossing | 221 | self-crossing | 0.293 | r0bertz | Hard | 5,790 | 335 |
palindrome pairs | class Solution:
def palindromePairs(self, words: List[str]) -> List[List[int]]:
backward, res = {}, []
for i, word in enumerate(words):
backward[word[::-1]] = i
for i, word in enumerate(words):
if word in backward and backward[word] != i:
... | https://leetcode.com/problems/palindrome-pairs/discuss/2585442/Intuitive-Python3-or-HashMap-or-95-Time-and-Space-or-O(N*W2) | 54 | You are given a 0-indexed array of unique strings words.
A palindrome pair is a pair of integers (i, j) such that:
0 <= i, j < words.length,
i != j, and
words[i] + words[j] (the concatenation of the two strings) is a
palindrome
.
Return an array of all the palindrome pairs of words.
You must write an algorithm with O(s... | Intuitive Python3 | HashMap | 95% Time & Space | O(N*W^2) | 3,000 | palindrome-pairs | 0.352 | ryangrayson | Hard | 5,791 | 336 |
house robber iii | class Solution:
def rob(self, root: TreeNode) -> int:
def dfs(node):
if not node: return 0, 0
left, right = dfs(node.left), dfs(node.right)
v_take = node.val + left[1] + right[1]
v_not_take = max(left) + max(right)
return v_take, v_not_take
... | https://leetcode.com/problems/house-robber-iii/discuss/872676/Python-3-or-DFS-Backtracking-or-Explanation | 5 | The thief has found himself a new place for his thievery again. There is only one entrance to this area, called root.
Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that all houses in this place form a binary tree. It will automatically contact the police if two d... | Python 3 | DFS, Backtracking | Explanation | 439 | house-robber-iii | 0.539 | idontknoooo | Medium | 5,801 | 337 |
counting bits | class Solution:
def countBits(self, N: int) -> List[int]:
stem = [0]
while len(stem) < N+1:
stem.extend([s + 1 for s in stem])
return stem[:N+1] | https://leetcode.com/problems/counting-bits/discuss/466438/Python-clean-no-cheat-easy-to-understand.-Based-on-pattern.-Beats-90. | 18 | Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1's in the binary representation of i.
Example 1:
Input: n = 2
Output: [0,1,1]
Explanation:
0 --> 0
1 --> 1
2 --> 10
Example 2:
Input: n = 5
Output: [0,1,1,2,1,2]
Explanation:
0 --> 0
1 --> 1
2 --> 10
3... | Python clean, no cheat, easy to understand. Based on pattern. Beats 90%. | 1,100 | counting-bits | 0.753 | kimonode | Easy | 5,817 | 338 |
power of four | class Solution:
def isPowerOfFour(self, num: int) -> bool:
return num > 0 and not num & (num - 1) and len(bin(num)) % 2 | https://leetcode.com/problems/power-of-four/discuss/772261/Python3-1-liner-99.95-O(1)-explained | 9 | Given an integer n, return true if it is a power of four. Otherwise, return false.
An integer n is a power of four, if there exists an integer x such that n == 4x.
Example 1:
Input: n = 16
Output: true
Example 2:
Input: n = 5
Output: false
Example 3:
Input: n = 1
Output: true
Constraints:
-231 <= n <= 231 - 1
Fol... | Python3 1 liner 99.95% O(1), explained | 546 | power-of-four | 0.458 | dangtrangiabao | Easy | 5,879 | 342 |
integer break | class Solution:
def integerBreak(self, n: int) -> int:
if(n<=3):
return n-1
n3=n//3
r3=n%3
if(r3==0):
return 3**n3
if(r3==1):
r3=4
n3-=1
return r3*(3**n3) | https://leetcode.com/problems/integer-break/discuss/2830343/O(1)-oror-TC1-10-line-code | 7 | Given an integer n, break it into the sum of k positive integers, where k >= 2, and maximize the product of those integers.
Return the maximum product you can get.
Example 1:
Input: n = 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.
Example 2:
Input: n = 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.
Co... | O(1) || TC=1 10 line code | 71 | integer-break | 0.554 | droj | Medium | 5,926 | 343 |
reverse string | class Solution:
def reverseString(self, s: List[str]) -> None:
s[:] = s[::-1] | https://leetcode.com/problems/reverse-string/discuss/670137/Python-3-~actually~-easiest-solution | 77 | Write a function that reverses a string. The input string is given as an array of characters s.
You must do this by modifying the input array in-place with O(1) extra memory.
Example 1:
Input: s = ["h","e","l","l","o"]
Output: ["o","l","l","e","h"]
Example 2:
Input: s = ["H","a","n","n","a","h"]
Output: ["h","a","n",... | Python 3 ~actually~ easiest solution | 6,800 | reverse-string | 0.762 | drblessing | Easy | 5,954 | 344 |
reverse vowels of a string | class Solution:
def reverseVowels(self, s: str) -> str:
s = list(s)
left = 0
right = len(s) - 1
m = 'aeiouAEIOU'
while left < right:
if s[left] in m and s[right] in m:
s[left], s[right] = s[right], s[left]
... | https://leetcode.com/problems/reverse-vowels-of-a-string/discuss/1164745/Python-Solution-oror-99.58-faster-oror-86.96-less-memory | 22 | Given a string s, reverse only all the vowels in the string and return it.
The vowels are 'a', 'e', 'i', 'o', and 'u', and they can appear in both lower and upper cases, more than once.
Example 1:
Input: s = "hello"
Output: "holle"
Example 2:
Input: s = "leetcode"
Output: "leotcede"
Constraints:
1 <= s.length <= 3 ... | Python Solution || 99.58% faster || 86.96% less memory | 1,100 | reverse-vowels-of-a-string | 0.498 | KiranUpase | Easy | 6,016 | 345 |
top k frequent elements | class Solution:
def topKFrequent(self, nums: List[int], k: int) -> List[int]:
frequency = {}
for num in nums:
if num not in frequency:
frequency[num] = 1
else:
frequency[num] = frequency[num] + 1
frequency = dict(sorted(frequency.items(), key=lambda x: x[1], reverse=True))
result = list(f... | https://leetcode.com/problems/top-k-frequent-elements/discuss/1928198/Python-Simple-Python-Solution-Using-Dictionary-(-HashMap-) | 24 | Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.
Example 1:
Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]
Example 2:
Input: nums = [1], k = 1
Output: [1]
Constraints:
1 <= nums.length <= 105
-104 <= nums[i] <= 104
k is in the range [1, the... | [ Python ] ✅✅ Simple Python Solution Using Dictionary ( HashMap ) ✌👍 | 2,800 | top-k-frequent-elements | 0.648 | ASHOK_KUMAR_MEGHVANSHI | Medium | 6,072 | 347 |
intersection of two arrays | class Solution:
def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
a = []
for i in nums1:
if i not in a and i in nums2:
a.append(i)
return a | https://leetcode.com/problems/intersection-of-two-arrays/discuss/2270388/PYTHON-3-SIMPLE-or-EASY-TO-UNDERSTAND | 5 | Given two integer arrays nums1 and nums2, return an array of their
intersection
. Each element in the result must be unique and you may return the result in any order.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]
Explanation: [4,9]... | [PYTHON 3] SIMPLE | EASY TO UNDERSTAND | 135 | intersection-of-two-arrays | 0.704 | omkarxpatel | Easy | 6,131 | 349 |
intersection of two arrays ii | class Solution:
def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
nums1.sort()
nums2.sort()
one=0
two=0
ans=[]
while one < len(nums1) and two < len(nums2):
if nums1[one] < nums2[... | https://leetcode.com/problems/intersection-of-two-arrays-ii/discuss/1231807/easy-or-two-pointer-method-or-python | 23 | Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,... | easy | two pointer method | python | 1,300 | intersection-of-two-arrays-ii | 0.556 | chikushen99 | Easy | 6,183 | 350 |
russian doll envelopes | class Solution:
def maxEnvelopes(self, envelopes: List[List[int]]) -> int:
envelopes.sort(key=lambda x: (x[0], -x[1]))
res = []
# Perform LIS
for _, h in envelopes:
l,r=0,len(res)-1
# find the insertion point in the Sort order
while l <= r:
... | https://leetcode.com/problems/russian-doll-envelopes/discuss/2071626/Python-LIS-based-approach | 19 | You are given a 2D array of integers envelopes where envelopes[i] = [wi, hi] represents the width and the height of an envelope.
One envelope can fit into another if and only if both the width and height of one envelope are greater than the other envelope's width and height.
Return the maximum number of envelopes you c... | Python LIS based approach | 1,800 | russian-doll-envelopes | 0.382 | constantine786 | Hard | 6,234 | 354 |
count numbers with unique digits | class Solution:
def countNumbersWithUniqueDigits(self, n: int) -> int:
if n == 0: return 1
if n == 1: return 10
res = 91
mult = 8
comb = 81
for i in range(n - 2):
comb *= mult
mult -= 1
res += comb
return res | https://leetcode.com/problems/count-numbers-with-unique-digits/discuss/2828071/Python3-Mathematics-approach.-Explained-in-details.-Step-by-step | 0 | Given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10n.
Example 1:
Input: n = 2
Output: 91
Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100, excluding 11,22,33,44,55,66,77,88,99
Example 2:
Input: n = 0
Output: 1
Constraints:
0 <= n <= 8 | Python3 Mathematics approach. Explained in details. Step-by-step | 2 | count-numbers-with-unique-digits | 0.516 | Alex_Gr | Medium | 6,239 | 357 |
max sum of rectangle no larger than k | class Solution:
def maxSumSubmatrix(self, matrix: List[List[int]], k: int) -> int:
ans = float("-inf")
m, n = len(matrix), len(matrix[0])
for i in range(n):
lstSum = [0] * m
for j in range(i, n):
currSum = 0
curlstSum = [0]
... | https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/discuss/2488882/Solution-In-Python | 8 | Given an m x n matrix matrix and an integer k, return the max sum of a rectangle in the matrix such that its sum is no larger than k.
It is guaranteed that there will be a rectangle with a sum no larger than k.
Example 1:
Input: matrix = [[1,0,1],[0,-2,3]], k = 2
Output: 2
Explanation: Because the sum of the blue rec... | Solution In Python | 1,200 | max-sum-of-rectangle-no-larger-than-k | 0.441 | AY_ | Hard | 6,250 | 363 |
water and jug problem | class Solution:
def canMeasureWater(self, x: int, y: int, z: int) -> bool:
return False if x + y < z else True if x + y == 0 else not z % math.gcd(x,y)
- Junaid Mansuri
(LeetCode ID)@hotmail.com | https://leetcode.com/problems/water-and-jug-problem/discuss/393886/Solution-in-Python-3-(beats-~100)-(one-line)-(Math-Solution) | 5 | You are given two jugs with capacities x liters and y liters. You have an infinite water supply. Return whether the total amount of water in both jugs may reach target using the following operations:
Fill either jug completely with water.
Completely empty either jug.
Pour water from one jug into another until the recei... | Solution in Python 3 (beats ~100%) (one line) (Math Solution) | 1,300 | water-and-jug-problem | 0.367 | junaidmansuri | Medium | 6,254 | 365 |
valid perfect square | class Solution:
def isPerfectSquare(self, num: int) -> bool:
return int(num**0.5) == num**0.5 | https://leetcode.com/problems/valid-perfect-square/discuss/1063963/100-Python-One-Liner-UPVOTE-PLEASE | 6 | Given a positive integer num, return true if num is a perfect square or false otherwise.
A perfect square is an integer that is the square of an integer. In other words, it is the product of some integer with itself.
You must not use any built-in library function, such as sqrt.
Example 1:
Input: num = 16
Output: true... | 100% Python One-Liner UPVOTE PLEASE | 451 | valid-perfect-square | 0.433 | 1coder | Easy | 6,265 | 367 |
largest divisible subset | class Solution:
def largestDivisibleSubset(self, nums: List[int]) -> List[int]:
if not nums or len(nums) == 0:
return []
# since we are doing a "subset" question
# sorting does not make any differences
nums.sort()
n = len(nums)
# initilization
# ... | https://leetcode.com/problems/largest-divisible-subset/discuss/1127633/Python-Dynamic-Programming-with-comments | 3 | Given a set of distinct positive integers nums, return the largest subset answer such that every pair (answer[i], answer[j]) of elements in this subset satisfies:
answer[i] % answer[j] == 0, or
answer[j] % answer[i] == 0
If there are multiple solutions, return any of them.
Example 1:
Input: nums = [1,2,3]
Output: [1,... | Python Dynamic Programming with comments | 227 | largest-divisible-subset | 0.413 | zna2 | Medium | 6,316 | 368 |
sum of two integers | class Solution:
def getSum(self, a: int, b: int) -> int:
return int(math.log2(2**a * 2**b)) | https://leetcode.com/problems/sum-of-two-integers/discuss/1876632/Python-one-line-solution-using-the-logic-of-logs-and-powers | 7 | Given two integers a and b, return the sum of the two integers without using the operators + and -.
Example 1:
Input: a = 1, b = 2
Output: 3
Example 2:
Input: a = 2, b = 3
Output: 5
Constraints:
-1000 <= a, b <= 1000 | Python one line solution using the logic of logs and powers | 596 | sum-of-two-integers | 0.507 | alishak1999 | Medium | 6,324 | 371 |
super pow | class Solution:
def superPow(self, a: int, b: List[int]) -> int:
return (a % 1337)**(1140 + int(''.join(map(str, b))) % 1140) % 1337
- Junaid Mansuri | https://leetcode.com/problems/super-pow/discuss/400893/Python-3-(With-Explanation)-(Handles-All-Test-Cases)-(one-line)-(beats-~97) | 11 | Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.
Example 1:
Input: a = 2, b = [3]
Output: 8
Example 2:
Input: a = 2, b = [1,0]
Output: 1024
Example 3:
Input: a = 1, b = [4,3,3,8,5,2]
Output: 1
Constraints:
1 <= a <= 231 -... | Python 3 (With Explanation) (Handles All Test Cases) (one line) (beats ~97%) | 1,900 | super-pow | 0.371 | junaidmansuri | Medium | 6,343 | 372 |
find k pairs with smallest sums | class Solution:
def kSmallestPairs(self, nums1: List[int], nums2: List[int], k: int) -> List[List[int]]:
hq = []
heapq.heapify(hq)
# add all the pairs that we can form with
# all the (first k) items in nums1 with the first
# item in nums2
for i in range(min(l... | https://leetcode.com/problems/find-k-pairs-with-smallest-sums/discuss/1701122/Python-Simple-heap-solution-explained | 11 | You are given two integer arrays nums1 and nums2 sorted in non-decreasing order and an integer k.
Define a pair (u, v) which consists of one element from the first array and one element from the second array.
Return the k pairs (u1, v1), (u2, v2), ..., (uk, vk) with the smallest sums.
Example 1:
Input: nums1 = [1,7,1... | [Python] Simple heap solution explained | 1,400 | find-k-pairs-with-smallest-sums | 0.383 | buccatini | Medium | 6,363 | 373 |
guess number higher or lower | class Solution:
def guessNumber(self, n: int) -> int:
return __pick__ | https://leetcode.com/problems/guess-number-higher-or-lower/discuss/2717871/DONT-TRY-THIS-CODE-or-ONE-LINE-PYTHON-CODE | 9 | We are playing the Guess Game. The game is as follows:
I pick a number from 1 to n. You have to guess which number I picked.
Every time you guess wrong, I will tell you whether the number I picked is higher or lower than your guess.
You call a pre-defined API int guess(int num), which returns three possible results:
-1... | DONT TRY THIS CODE | ONE LINE PYTHON CODE | 441 | guess-number-higher-or-lower | 0.514 | raghavdabra | Easy | 6,377 | 374 |
guess number higher or lower ii | class Solution:
def getMoneyAmount(self, n: int) -> int:
if n == 1:
return 1
starting_index = 1 if n % 2 == 0 else 2
selected_nums = [i for i in range(starting_index, n, 2)]
selected_nums_length = len(selected_nums)
dp = [[0] * selected_nums_length for _ in range(... | https://leetcode.com/problems/guess-number-higher-or-lower-ii/discuss/1510747/Python-DP-beat-97.52-in-time-99-in-memory-(with-explanation) | 31 | We are playing the Guessing Game. The game will work as follows:
I pick a number between 1 and n.
You guess a number.
If you guess the right number, you win the game.
If you guess the wrong number, then I will tell you whether the number I picked is higher or lower, and you will continue guessing.
Every time you guess ... | [Python] DP, beat 97.52% in time, 99% in memory (with explanation) | 1,100 | guess-number-higher-or-lower-ii | 0.465 | wingskh | Medium | 6,399 | 375 |
wiggle subsequence | class Solution:
def wiggleMaxLength(self, nums: List[int]) -> int:
length = 0
curr = 0
for i in range(len(nums) - 1):
if curr == 0 and nums[i + 1] - nums[i] != 0:
length += 1
curr = nums[i + 1] - nums[i]
if cur... | https://leetcode.com/problems/wiggle-subsequence/discuss/2230152/Beats-73.3-Simple-Python-Solution-Greedy | 3 | A wiggle sequence is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.
For example, ... | Beats 73.3% - Simple Python Solution - Greedy | 183 | wiggle-subsequence | 0.482 | 7yler | Medium | 6,406 | 376 |
combination sum iv | class Solution:
def combinationSum4(self, nums: List[int], target: int) -> int:
waysToAdd = [0 for x in range(target+1)]
waysToAdd[0] = 1
for i in range(min(nums), target+1):
waysToAdd[i] = sum(waysToAdd[i-num] for num in nums if i-num >= 0)
return waysT... | https://leetcode.com/problems/combination-sum-iv/discuss/1272869/Python-3-Faster-than-96-(Super-Simple!) | 3 | Given an array of distinct integers nums and a target integer target, return the number of possible combinations that add up to target.
The test cases are generated so that the answer can fit in a 32-bit integer.
Example 1:
Input: nums = [1,2,3], target = 4
Output: 7
Explanation:
The possible combination ways are:
(1... | [Python 3] Faster than 96% (Super Simple!) | 323 | combination-sum-iv | 0.521 | jodoko | Medium | 6,438 | 377 |
kth smallest element in a sorted matrix | class Solution:
def kthSmallest(self, matrix: List[List[int]], k: int) -> int:
m = len(matrix)
n = len(matrix[0])
def count(m):
c = 0 # count of element less than equals to 'm'
i = n-1
j = 0
whil... | https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/discuss/2233868/Simple-yet-best-Interview-Code-or-Python-Code-beats-90 | 5 | Given an n x n matrix where each of the rows and columns is sorted in ascending order, return the kth smallest element in the matrix.
Note that it is the kth smallest element in the sorted order, not the kth distinct element.
You must find a solution with a memory complexity better than O(n2).
Example 1:
Input: matri... | ✅ Simple yet best Interview Code | Python Code beats 90% | 532 | kth-smallest-element-in-a-sorted-matrix | 0.616 | reinkarnation | Medium | 6,467 | 378 |
linked list random node | class Solution:
def __init__(self, head: ListNode):
"""
@param head The linked list's head.
Note that the head is guaranteed to be not null, so it contains at least one node.
"""
self.head = head # store head of linked list
def getRandom(self) -> int:
"""
... | https://leetcode.com/problems/linked-list-random-node/discuss/811617/Python3-reservoir-sampling | 4 | Given a singly linked list, return a random node's value from the linked list. Each node must have the same probability of being chosen.
Implement the Solution class:
Solution(ListNode head) Initializes the object with the head of the singly-linked list head.
int getRandom() Chooses a node randomly from the list and re... | [Python3] reservoir sampling | 395 | linked-list-random-node | 0.596 | ye15 | Medium | 6,513 | 382 |
ransom note | class Solution:
def canConstruct(self, ransomNote, magazine):
for i in set(ransomNote):
if magazine.count(i) < ransomNote.count(i):
return False
return True | https://leetcode.com/problems/ransom-note/discuss/1346131/Easiest-python-solution-faster-than-95 | 35 | Given two strings ransomNote and magazine, return true if ransomNote can be constructed by using the letters from magazine and false otherwise.
Each letter in magazine can only be used once in ransomNote.
Example 1:
Input: ransomNote = "a", magazine = "b"
Output: false
Example 2:
Input: ransomNote = "aa", magazine = ... | Easiest python solution, faster than 95% | 3,500 | ransom-note | 0.576 | mqueue | Easy | 6,520 | 383 |
shuffle an array | class Solution:
def __init__(self, nums: List[int]):
self.arr = nums[:] # Deep Copy, Can also use Shallow Copy concept!
# self.arr = nums # Shallow Copy would be something like this!
def reset(self) -> List[int]:
return self.arr
def shuffle(self) -> List[int]:
ans = self.arr[:]
for i in range(len(ans))... | https://leetcode.com/problems/shuffle-an-array/discuss/1673643/Python-or-Best-Optimal-Approach | 6 | Given an integer array nums, design an algorithm to randomly shuffle the array. All permutations of the array should be equally likely as a result of the shuffling.
Implement the Solution class:
Solution(int[] nums) Initializes the object with the integer array nums.
int[] reset() Resets the array to its original confi... | { Python } | Best Optimal Approach ✔ | 725 | shuffle-an-array | 0.577 | leet_satyam | Medium | 6,569 | 384 |
mini parser | class Solution:
def deserialize(self, s: str) -> NestedInteger:
if not s: return NestedInteger()
if not s.startswith("["): return NestedInteger(int(s)) # integer
ans = NestedInteger()
s = s[1:-1] # strip outer "[" and "]"
if s:
ii = op = 0
for i in ... | https://leetcode.com/problems/mini-parser/discuss/875743/Python3-a-concise-recursive-solution | 2 | Given a string s represents the serialization of a nested list, implement a parser to deserialize it and return the deserialized NestedInteger.
Each element is either an integer or a list whose elements may also be integers or other lists.
Example 1:
Input: s = "324"
Output: 324
Explanation: You should return a Neste... | [Python3] a concise recursive solution | 178 | mini-parser | 0.366 | ye15 | Medium | 6,590 | 385 |
lexicographical numbers | class Solution:
def lexicalOrder(self, n: int) -> List[int]:
return sorted([x for x in range(1,n+1)],key=lambda x: str(x)) | https://leetcode.com/problems/lexicographical-numbers/discuss/2053392/Python-oneliner | 1 | Given an integer n, return all the numbers in the range [1, n] sorted in lexicographical order.
You must write an algorithm that runs in O(n) time and uses O(1) extra space.
Example 1:
Input: n = 13
Output: [1,10,11,12,13,2,3,4,5,6,7,8,9]
Example 2:
Input: n = 2
Output: [1,2]
Constraints:
1 <= n <= 5 * 104 | Python oneliner | 135 | lexicographical-numbers | 0.608 | StikS32 | Medium | 6,592 | 386 |
first unique character in a string | class Solution:
def firstUniqChar(self, s: str) -> int:
for i in range(len(s)):
if s[i] not in s[:i] and s[i] not in s[i+1:]:
return i
return -1 | https://leetcode.com/problems/first-unique-character-in-a-string/discuss/1793386/Python-Simple-Python-Solution-With-Two-Approach | 28 | Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.
Example 1:
Input: s = "leetcode"
Output: 0
Example 2:
Input: s = "loveleetcode"
Output: 2
Example 3:
Input: s = "aabb"
Output: -1
Constraints:
1 <= s.length <= 105
s consists of only lowercase Engli... | [ Python ] ✅✅ Simple Python Solution With Two Approach 🥳✌👍 | 1,600 | first-unique-character-in-a-string | 0.59 | ASHOK_KUMAR_MEGHVANSHI | Easy | 6,607 | 387 |
longest absolute file path | class Solution:
def lengthLongestPath(self, s: str) -> int:
paths, stack, ans = s.split('\n'), [], 0
for path in paths:
p = path.split('\t')
depth, name = len(p) - 1, p[-1]
l = len(name)
while stack and stack[-1][1] >= depth: stack.pop()
if... | https://leetcode.com/problems/longest-absolute-file-path/discuss/812407/Python-3-or-Stack-or-Explanation | 23 | Suppose we have a file system that stores both files and directories. An example of one system is represented in the following picture:
Here, we have dir as the only directory in the root. dir contains two subdirectories, subdir1 and subdir2. subdir1 contains a file file1.ext and subdirectory subsubdir1. subdir2 contai... | Python 3 | Stack | Explanation | 1,900 | longest-absolute-file-path | 0.465 | idontknoooo | Medium | 6,655 | 388 |
find the difference | class Solution:
def findTheDifference(self, s: str, t: str) -> str:
s, t = sorted(s), sorted(t)
for i,j in enumerate(s):
if j != t[i]: return t[i]
return t[-1] | https://leetcode.com/problems/find-the-difference/discuss/379846/Three-Solutions-in-Python-3 | 23 | You are given two strings s and t.
String t is generated by random shuffling string s and then add one more letter at a random position.
Return the letter that was added to t.
Example 1:
Input: s = "abcd", t = "abcde"
Output: "e"
Explanation: 'e' is the letter that was added.
Example 2:
Input: s = "", t = "y"
Output:... | Three Solutions in Python 3 | 2,300 | find-the-difference | 0.603 | junaidmansuri | Easy | 6,673 | 389 |
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