Number of Even and Odd Bits

Problem Statement - link #

You are given a positive integer n.

Let even denote the number of even indices in the binary representation of n (0-indexed) with value 1.

Let odd denote the number of odd indices in the binary representation of n (0-indexed) with value 1.

Return an integer array answer where answer = [even, odd].

Examples #

Example 1:

Input: n = 17
Output: [2,0]
Explanation: The binary representation of 17 is 10001. 
It contains 1 on the 0th and 4th indices. 
There are 2 even and 0 odd indices.

Example 2:

Input: n = 2
Output: [0,1]
Explanation: The binary representation of 2 is 10.
It contains 1 on the 1st index. 
There are 0 even and 1 odd indices.

Constraints #

• 1 <= n <= 1000

Solutions #

class Solution {
public:
    vector<int> evenOddBit(int n) {
        int id = 0;
        int even = 0, odd = 0;
        while(id < 12) {
            if(n & (1 << id)) {
                even++;
            }
            id++;
            if(n & (1 << id)) {
                odd++;
            }
            id++;
        }
        return {even, odd};
    }
};