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XOR counts of 0s and 1s in binary representation
  • Difficulty Level : Easy
  • Last Updated : 14 Apr, 2021

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

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
 



This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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