LeetCode-in-Java
2595. Number of Even and Odd Bits
Easy
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
Solution
public class Solution {
public int[] evenOddBit(int n) {
int[] res = new int[2];
int even = 0;
int odd = 0;
for (int i = 0; i < 32; i++) {
if (i % 2 == 0) {
if ((n & 1) == 1) {
even++;
}
} else {
if ((n & 1) == 1) {
odd++;
}
}
n >>= 1;
}
res[0] = even;
res[1] = odd;
return res;
}
}