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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 :

Leetcode 338. Counting Bits. Python (Bubble Sort)

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 --> 11 4 --> 100 5 --> 101 Constraints: 0 <= n <= 10 5 Follow up: It is very easy to come up with a solution with a runtime of O(n log n) . Can you do it in linear time O(n) and possibly in a single pass? Can you do it without using any built-in function (i.e., like __builtin_popcount in C++)? Solution : class Solution: def countBits(self, n: int) -> List[int]: dp = [0] * (n+1) c = 1 ...

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...

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