POTD Solutions | 30 Oct’ 23 | Sum of XOR of all pairs

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Welcome to the daily solutions of our PROBLEM OF THE DAY (POTD). We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Bit Manipulation but will also help you build up problem-solving skills.

POTD 30 Oct’ 2023

POTD 30 October: Sum of XOR of all pairs

Given an array of N integers, find the sum of xor of all pairs of numbers in the array arr. In other words, select all possible pairs of i and j from 0 to N-1 (i<j) and determine sum of all (arr_i xor arr_j).

Example 1:
Input: arr[] = {7, 3, 5}
Output: 12
Explanation: All possible pairs and sum of their XOR Values: (3 ^ 5) + (7 ^ 3) + (7 ^ 5) = 6 + 4 + 2 = 12

Example 2:
Input: {5, 9, 7, 6}
Output: 47
Explanation: All possible pairs and sum of their XOR Values: (5 ^ 9) + (5 ^ 7) + (5 ^ 6) + (9 ^ 7) + (9 ^ 6) + (7 ^ 6) = 12 + 2 + 3 + 14 + 15 + 1 = 47

Sum of XOR of all pairs using Bit Manipulation:

The idea is that for every bit position i, we consider all bits which are 1 and which are 0 and store their count in two different variables. Next multiply those counts along with the power of 2 raised to that bit position i. Do this for all the bit positions of the numbers. Their sum would be our answer.

For any bit position i, suppose A numbers have their ith bit unset (ith bit = 0) and B numbers have their ith bit set (ith bit = 1). Then out of all the pairs, A*B pairs will have the ith bit set in their XOR. This is because there are A*B ways to choose one number that has a 0-bit and one that has a 1-bit. These bits will therefore contribute A*B towards the total of all the XORs. The contribution towards the final sum will be A*B*pow(2,i). Do this for each bit and sum all these contributions together.

Below is the implementation of the above approach:

C++

class Solution {
public:
    // Returns sum of bitwise OR
    // of all pairs
    long long int sumXOR(int arr[], int n)
    {
        // Complete the function
        long long int sum = 0;
        for (int i = 0; i < 32; i++) {
            //  Count of zeros and ones
            int A = 0, B = 0;

            // Individual sum at each bit position
            long long int idsum = 0;
            for (int j = 0; j < n; j++) {
                if (arr[j] % 2 == 0)
                    A++;
                else
                    B++;
                arr[j] /= 2;
            }

            // calculating individual bit sum
            idsum = A * 1LL * B * (1LL << i);

            // final sum
            sum += idsum;
        }
        return sum;
    }
};

Java

class Solution {

    // Function for finding maximum and value pair
    public long sumXOR(int arr[], int n)
    {
        // Complete the function

        long sum = 0;
        for (int i = 0; i < 32; i++) {
            // Count of zeros and ones
            long A = 0, B = 0;

            // Individual sum at each bit position
            long idsum = 0;

            for (int j = 0; j < n; j++) {
                if (arr[j] % 2 == 0)
                    A++;

                else
                    B++;
                arr[j] /= 2;
            }

            // calculating individual bit sum
            idsum = A * B * (1 << i);

            // final sum
            sum += idsum;
        }
        return sum;
    }
}

Python3

class Solution:
    def sumXOR(self, arr, n):
        sum = 0
        for i in range(0, 32):

            #  Count of zeros and ones
            A = 0
            B = 0

            # Individual sum at each bit position
            idsum = 0
            for j in range(0, n):
                if (arr[j] % 2 == 0):
                    A = A + 1

                else:
                    B = B + 1
                arr[j] = arr[j] // 2

            # calculating individual bit sum
            idsum = A * B * (1 << i)

            # final sum
            sum = sum + idsum

        return sum

Time Complexity: O(K * N), where K is the number of bits in the input array and N is the size of array
Auxiliary Space: O(1)