XOR counts of 0s and 1s in binary representation

Given a number, the task is to find XOR of count of 0s and count of 1s in binary representation of a given number. 
Examples:

Input  : 5
Output : 3
Binary representation : 101
Count of 0s = 1, 
Count of 1s = 2
1 XOR 2 = 3.

Input  : 7
Output : 3
Binary representation : 111
Count of 0s = 0
Count of 1s = 3
0 XOR 3 = 3.

The idea is simple, we traverse through all bits of a number, count 0s and 1s and finally return XOR of two counts.

C++

// C++ program to find XOR of counts 0s and 1s in
// binary representation of n.
#include<iostream>
using namespace std;

// Returns XOR of counts 0s and 1s in
// binary representation of n.
int countXOR(int n)
{
    int count0 = 0, count1 = 0;
    while (n)
    {
        //calculating count of zeros and ones
        (n % 2 == 0) ? count0++ :count1++;
        n /= 2;
    }
    return (count0 ^ count1);
}

// Driver Program
int main()
{
    int n = 31;
    cout << countXOR (n);
    return 0;
}

Java

// Java program to find XOR of counts 0s
// and 1s in binary representation of n.

class GFG {

    // Returns XOR of counts 0s and 1s
    // in binary representation of n.
    static int countXOR(int n)
    {
        int count0 = 0, count1 = 0;
        while (n != 0)
        {
            //calculating count of zeros and ones
            if(n % 2 == 0)
            count0++ ;
            else
            count1++;
            n /= 2;
        }
        return (count0 ^ count1);
    }

    // Driver Program
    public static void main(String[] args)
    {
        int n = 31;
        System.out.println(countXOR (n));
    }
}

// This code is contributed by prerna saini

Python3

# Python3 program to find XOR of counts 0s
# and 1s in binary representation of n.

# Returns XOR of counts 0s and 1s
# in binary representation of n.
def countXOR(n):

    count0, count1 = 0, 0
    while (n != 0):

        # calculating count of zeros and ones
        if(n % 2 == 0):
            count0 += 1
        else:
            count1 += 1
        n //= 2

    return (count0 ^ count1)

# Driver Code
n = 31
print(countXOR(n))

# This code is contributed by Anant Agarwal.

C#

// C# program to find XOR of counts 0s
// and 1s in binary representation of n.
using System;

class GFG {

    // Returns XOR of counts 0s and 1s
    // in binary representation of n.
    static int countXOR(int n)
    {
        int count0 = 0, count1 = 0;
        while (n != 0)
        {

            // calculating count of zeros
            // and ones
            if(n % 2 == 0)
                count0++ ;
            else
                count1++;

            n /= 2;
        }

        return (count0 ^ count1);
    }

    // Driver Program
    public static void Main()
    {

        int n = 31;

        Console.WriteLine(countXOR (n));
    }
}

// This code is contributed by Anant Agarwal.

PHP

<?PHP
// PHP program to find XOR of
// counts 0s and 1s in binary
// representation of n.

// Returns XOR of counts 0s and 1s
// in binary representation of n.
function countXOR($n)
{
    $count0 = 0;
    $count1 = 0;
    while ($n)
    {
        // calculating count of
        // zeros and ones
        ($n % 2 == 0) ? $count0++ :$count1++;
        $n = intval($n / 2);
    }
    return ($count0 ^ $count1);
}

// Driver Code
$n = 31;
echo countXOR ($n);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>

// Javascript program to find XOR of counts 0s
// and 1s in binary representation of n.

    // Returns XOR of counts 0s and 1s
    // in binary representation of n.
    function  countXOR(n)
    {
        let count0 = 0, count1 = 0;
        while (n != 0)
        {
            //calculating count of zeros and ones
            if(n % 2 == 0)
            count0++ ;
            else
            count1++;
            n = Math.floor(n/2);
        }
        return (count0 ^ count1);
    }

    // Driver Program
    let n = 31;
    document.write(countXOR (n));

    // This code is contributed by avanitrachhadiya2155

</script>

Output:

5

Time Complexity: O(log(N))
Auxiliary Space: O(1)

One observation is, for a number of the form 2^x – 1, the output is always x. We can directly produce answer for this case by first checking n+1 is a power of two or not.