Number of Even and Odd Bits

 You are given a positive integer n.


Let even denote the number of even indices in the binary representation of n with value 1.


Let odd denote the number of odd indices in the binary representation of n with value 1.


Note that bits are indexed from right to left in the binary representation of a number.


Return the array [even, odd].


 


Example 1:


Input: n = 50


Output: [1,2]


Explanation:


The binary representation of 50 is 110010.


It contains 1 on indices 1, 4, and 5.


Example 2:


Input: n = 2


Output: [0,1]


Explanation:


The binary representation of 2 is 10.


It contains 1 only on index 1.


 


Constraints:


1 <= n <= 1000



class Solution {
    public int[] evenOddBit(int n) {
        int[] result = new int[2]; // result[0] for even, result[1] for odd
        int position = 0;

        while (n > 0) {
            if ((n & 1) == 1) {
                if (position % 2 == 0) {
                    result[0]++; // even position
                } else {
                    result[1]++; // odd position
                }
            }
            n >>= 1; // right shift to check next bit
            position++;
        }

        return result;
    }
}

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