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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 :
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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        ...
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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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