def sum_xor(n):
    # Time complexity: O(log(n))
    # Space complexity (ignoring input): O(1)
    if n == 0:
        return 1
    # count 0 in binary of number
    count = 0
    while n > 0:
        if n & 1 == 0:
            count += 1
        n = n >> 1

    return 2**count

fn sum_xor(n: i64) -> i64 {
    // Time complexity: O(log(n))
    // Space complexity (ignoring input): O(1)
    let mut n = n;
    let mut zeros_in_binary = 0;
    while n > 1 {
        if n & 1 == 0 {
            zeros_in_binary += 1;
        }
        n = n >> 1;
    }

    // The answer is that the number can have 1 or 0 in the positions where 'n' has 0, so 2 to the n
    return 1<<zeros_in_binary
}

public static long sumXor(long n) {
// Write your code here
Long res = 0L, nn = n;
res = (long)Math.pow(2,(Long.toBinaryString(nn).chars().filter(x -> x == '0').count()));
return n == 0 ? 1 : res;
}

fn sumXor(n: i64) -> i64 {
    let zeros = n.count_zeros() - n.leading_zeros();
    2_i64.pow(zeros)
}