Sum of Bitwise And of all pairs in a given array - GeeksforGeeks

Sum of Bitwise And of all pairs in a given array - GeeksforGeeks
Given an array "arr[0..n-1]" of integers, calculate sum of "arr[i] & arr[j]" for all the pairs in the given where i < j. Here & is bitwise AND operator. Expected time complexity is O(n).

Input:  arr[] = {5, 10, 15}
Output: 15
Required Value = (5 & 10) + (5 & 15) + (10 & 15) 
               = 0 + 5 + 10 
               = 15

Input: arr[] = {1, 2, 3, 4}
Output: 9
Required Value = (1 & 2) + (1 & 3) + (1 & 4) + 
                 (2 & 3) + (2 & 4) + (3 & 4) 
               = 0 + 1 + 0 + 2 + 0 + 0
               = 3

An Efficient Solution can solve this problem in O(n) time. The assumption here is that integers are represented using 32 bits.

The idea is to count number of set bits at every i’th position (i>=0 && i<=31). Any i'th bit of the AND of two numbers is 1 iff the corresponding bit in both the numbers is equal to 1. Let k be the count of set bits at i'th position. Total number of pairs with i'th set bit would be kC2 = k*(k-1)/2 (Count k means there are k numbers which have i’th set bit). Every such pair adds 2i to total sum. Similarly, we work for all other places and add the sum to our final answer.

A Brute Force approach is to run two loops and time complexity is O(n²).

// Returns value of "arr[0] & arr[1] + arr[0] & arr[2] +

// ... arr[i] & arr[j] + ..... arr[n-2] & arr[n-1]"

int pairAndSum(int arr[], int n)

{

    int ans = 0;  // Initialize result

 

    // Consider all pairs (arr[i], arr[j) such that

    // i < j

    for (int i = 0; i < n; i++)

        for (int j = i+1; j < n; j++)

           ans += arr[i] & arr[j];

 

    return ans;

}

This idea is similar to Sum of bit differences among all pairs
Read full article from Sum of Bitwise And of all pairs in a given array - GeeksforGeeks