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Formatted question description: https://leetcode.ca/all/2595.html

2595. Number of Even and Odd Bits

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

Solution 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.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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;
      }
  }
  

• 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;
      }
  };
  

• 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
  
  

• 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
  }
  

• 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;
  }
  
  

• impl Solution {
      pub fn even_odd_bit(mut n: i32) -> Vec<i32> {
          let mut ans = vec![0; 2];
  
          let mut i = 0;
          while n != 0 {
              ans[i] += n & 1;
  
              n >>= 1;
              i ^= 1;
          }
  
          ans
      }
  }
  

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