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++ 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; }

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.

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