Sum of XOR of all pairs in an array in C++

In this problem, we are given an array arr[] of n integers. Our task is to create a program to find the sum of XOR of all pairs in an array.

Let’s take an example to understand the problem,

Input: arr[] = {5, 1, 4}
Output: 10
Explanation: the sum of all pairs:
5 ^ 1 = 4
1 ^ 4 = 5
5 ^ 4 = 1
sum = 4 + 5 + 1 = 10

One simple approach to solve this problem is to run nested loops and find all pairs of numbers. Find XOR of each pair and add them to the sum.

Algorithm

Initialise sum = 0
Step 1: for(i -> 0 to n). Do:
   Step 1.1: for( j -> i to n ). Do:
      Step 1.1.1: update sum. i.e. sum += arr[i] ^ arr[j].
Step 2: return sum.

Example

Program to illustrate the working of our solution,

 Live Demo

#include <iostream>
using namespace std;
int findXORSum(int arr[], int n) {
   int sum = 0;
   for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
    sum += (arr[i]^arr[j]);
   return sum;
}
int main() {
   int arr[] = { 5, 1, 4 };
   int n = sizeof(arr) / sizeof(arr[0]);
   cout<<"Sum of XOR of all pairs in an array is "<<findXORSum(arr, n);
   return 0;
}

Output

Sum of XOR of all pairs in an array is 10

The time complexity of the algorithm is O(n2) and is not the most efficient solution to the problem.

An efficient solution to the problem is using the bit manipulation technique.

Here, we will consider bits of the number and at each position. And apply the below formula find the intermediate sum,

(number of set bits) * (number of unset bits) * (2^(bit_position))

To find the final sum, we will add the intermediate sum of all bits.

Our solution is for the 64-bit integer value. For this approach, we need the number of bits.

Algorithm

Initialize sum = 0, setBits = 0, unsetBits = 0.
Step 1: Loop for i -> 0 to 64. repeat steps 2, 3.
Step 2: Reset setBits and unsetBits to 0.
Step 3: For each element of the array find the value of setBits and unsetBits. At ith position.
Step 4: update sum += (setBits * unsetBits * (2i))

Example

Program to illustrate the working of our solution,

 Live Demo

#include <iostream>
#include <math.h>
using namespace std;
long findXORSum(int arr[], int n) {
   long sum = 0;
   int unsetBits = 0, setBits = 0;
   for (int i = 0; i < 32; i++) {
      unsetBits = 0; setBits = 0;
      for (int j = 0; j < n; j++) {
         if (arr[j] % 2 == 0)
            unsetBits++;
         else
            setBits++;
         arr[j] /= 2;
      }
      sum += ( unsetBits*setBits* (pow(2,i)) );
   }
   return sum;
}
int main() {
   int arr[] = { 5, 1, 4, 7, 9};
   int n = sizeof(arr) / sizeof(arr[0]);
   cout<<"Sum of XOR of all pairs in an array is "<<findXORSum(arr, n);
   return 0;
}

Output

Sum of XOR of all pairs in an array is 68