README_EN.md

2595. Number of Even and Odd Bits

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Description

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].

 

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

Approach 1: Enumerate

According to the problem description, enumerate the binary representation of $n$ from the low bit to the high bit. If the bit is $1$, add $1$ to the corresponding counter according to whether the index of the bit is odd or even.

The time complexity is $O(\log n)$ and the space complexity is $O(1)$. Where $n$ is the given integer.

Python3

class Solution:
    def evenOddBit(self, n: int) -> List[int]:
        ans = [0, 0]
        i = 0
        while n:
            ans[i] += n & 1
            i ^= 1
            n >>= 1
        return ans

Java

class Solution {
    public int[] evenOddBit(int n) {
        int[] ans = new int[2];
        for (int i = 0; n > 0; n >>= 1, i ^= 1) {
            ans[i] += n & 1;
        }
        return ans;
    }
}

C++

class Solution {
public:
    vector<int> evenOddBit(int n) {
        vector<int> ans(2);
        for (int i = 0; n > 0; n >>= 1, i ^= 1) {
            ans[i] += n & 1;
        }
        return ans;
    }
};

Go

func evenOddBit(n int) []int {
	ans := make([]int, 2)
	for i := 0; n != 0; n, i = n>>1, i^1 {
		ans[i] += n & 1
	}
	return ans
}

TypeScript

function evenOddBit(n: number): number[] {
    const ans = new Array(2).fill(0);
    for (let i = 0; n > 0; n >>= 1, i ^= 1) {
        ans[i] += n & 1;
    }
    return ans;
}

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