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.

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



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.

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