# Hackerrank Sum vs XoR

I'm in the process of improving my coding style and performance hence I decided to delve into bit manipulation on Hackerrank. I thought the question was straight forward but on large input, my solution's performance declines. How do I improve it? Any smart tricks to solve the below question is welcome as well.

Below is the question

Given an integer, $n$ , find each $x$ such that:

$0 \leq x \leq n$

$n + x = n \oplus x$

where $\oplus$ denotes the bitwise XOR operator. Then print an integer denoting the total number of $x$ 's satisfying the criteria above.

## Input Format

A single integer, $n$ .

## Constraints

$0 \leq n \leq 10 ^ {15}$

$0 \leq n \leq 100$ for $60\%$ of the maximum score

## Output Format

Print the total number of integer $x$ 's satisfying both of the conditions specified above.

Sample Input 0

5


Sample Output 0

2


Explanation 0

For $n = 5$ , the $x$ values $0$ and $2$ satisfy the conditions:

Thus, we print $2$ as our answer.

Here is my code

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
class Solution {
static int SumVsXoR(long number)
{
int count =0;
long j = 0;
while (j <= number)
{
if(j + number == (j^number)){
count++;
}
j++;
}
return count;
}

static void Main(String[] args) {
Console.WriteLine(SumVsXoR(n));
}
}


Let's notice that
$n + x = n \oplus x$ is true
when and only when
$n + x = n \vee x$ is true

This is because $n + x \geq n \vee x = n \oplus x + n \wedge x$.

Where:

• $\vee$ - bitwise OR
• $\wedge$ - bitwise AND

So we just need to calculate the number of zero bits in the $n$.
Let's calculate the nonzero bits:

private static int NumberOfSetBits(ulong i)
{
i = i - ((i >> 1) & 0x5555555555555555UL);
i = (i & 0x3333333333333333UL) + ((i >> 2) & 0x3333333333333333UL);
return (int)(unchecked(((i + (i >> 4)) & 0xF0F0F0F0F0F0F0FUL) * 0x101010101010101UL) >> 56);
}


The method above was copy-pasted from this answer: stackoverflow.com

Then rewrite the SumVsXoR method:

private static int SumVsXoR(ulong number)
{
int numberOfBits = (int)Math.Log(number, 2) + 1;
return 1 << (numberOfBits - NumberOfSetBits(number));
}