Leetcode 226. Invert Binary Tree. Python (Very Easy)

226. Invert Binary Tree

class Solution:

def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:

if not root:

return None

temp = root.left

root.left = root.right

root.right = temp

self.invertTree(root.left)

self.invertTree(root.right)

return root

Explaination :

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Leetcode 371. Sum of Two Integers. C++ / Java

371 . Sum of Two Integers 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 Solution : C++ : class Solution { public: int getSum(int a, int b) { if (b==0) return a; int sum = a ^ b; int cr = (unsigned int) (a & b) << 1; return getSum(sum, cr); } }; Java : class Solution { public int getSum(int a, int b) { while(b != 0){ int tmp = (a & b) << 1; a = a ^ b; b = tmp; } return a; } } Explaination :

Number of Connected Components in an Undirected Graph (Python)

66. Number of Connected Components in an Undirected Graph Question Link : check here Givennnodes labeled from0ton - 1and a list of undirected edges (each edge is a pair of nodes), write a function to find the number of connected components in an undirected graph. Example 1: 0 3 | | 1 --- 2 4 Givenn = 5andedges = [[0, 1], [1, 2], [3, 4]], return2. Example 2: 0 4 | | 1 --- 2 --- 3 Givenn = 5andedges = [[0, 1], [1, 2], [2, 3], [3, 4]], return1. Note: You can assume that no duplicate edges will appear inedges. Since all edges are undirected,[0, 1]is the same as[1, 0]and thus will not appear together inedges. Solution : class Solution: def counComponents(self, n: int, edges : List[List[int]]) -> i...

Leetcode 347. Top K Frequent Elements. Python (Bubble Sort)

Top K Frequent Elements 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 <= 10 5 -10 4 <= nums[i] <= 10 4 k is in the range [1, the number of unique elements in the array] . It is guaranteed that the answer is unique . Follow up: Your algorithm's time complexity must be better than O(n log n) , where n is the array's size. Solution : class Solution: def topKFrequent(self, n: List[int], k: int) -> List[int]: # [1,1,1,2,2,3] & k = 2 f = [[] for i in range(len(n) + 1)] # f = [...

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