Java solution - passes 100% of test cases

From my HackerRank solutions.

XOR represents binary addition without the "carry" for each digit.

We want to see how many values of x that will give us n+x = n^x, which is when XOR is the same as ADDITION. This happens when there are no "carries". To make sure there are no carries, for each digit in "n" that is a 1, we must have the corresponding digit in "x" be a 0. For each digit in "n" that is a 0, we can have the corresponding digit in "x" be either 0 or 1.

Now we calculate the number of possibilities for "x" by counting the number of 0s in "n " (up to the most significant 1 in n), and doing 2^(number of 0s) (where ^ is exponentiation in this case) to count all combinations.

public class Solution {
    public static void main(String [] args) {
        /* Save input */
        Scanner scan = new Scanner(System.in);
        long n = scan.nextLong();
        scan.close();

        /* Calculate number of "x" values that will satisfy n+x = n^x */
        int zeroes = numZeroes(n);
        long result = 1L << zeroes; // same as (long) Math.pow(2, zeroes);
        System.out.println(result);
    }
    
    private static int numZeroes(long n) {
        int count = 0;
        while (n > 0) {
            if ((n & 1) == 0) {
                count++;
            }
            n >>= 1; // divides by 2
        }
        return count;
    }
}

Let me know if you have any questions.