Sum of Bitwise And of all pairs in a given array

3.6

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).
Examples:

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: 3
Required Value = (1 & 2) + (1 & 3) + (1 & 4) + 
                 (2 & 3) + (2 & 4) + (3 & 4) 
               = 0 + 1 + 0 + 2 + 0 + 0
               = 3

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

// A Simple C++ program to compute sum of bitwise AND 
// of all pairs
#include <bits/stdc++.h>
using namespace std;

// 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;
}

// Driver program to test above function
int main()
{
    int arr[] = {5, 10, 15};
    int n = sizeof(arr) / sizeof (arr[0]);
    cout << pairAndSum(arr, n) << endl;
    return 0;
}

Output:

15

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.

This idea is similar to this. Below is C++ implementation.

// An efficient C++ program to compute sum of bitwise AND
// of all pairs
#include <bits/stdc++.h>
using namespace std;

// 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

    // Traverse over all bits
    for (int i = 0; i < 32; i++)
    {
        // Count number of elements with i'th bit set
        int k = 0;  // Initialize the count
        for (int j = 0; j < n; j++)
            if ( (arr[j] & (1 << i)) )
                k++;

        // There are k set bits, means k(k-1)/2 pairs.
        // Every pair adds 2^i to the answer. Therefore,
        // we add "2^i * [k*(k-1)/2]" to the answer.
        ans += (1<<i) * (k*(k-1)/2);
    }

    return ans;
}

// Driver program to test above function
int main()
{
    int arr[] = {5, 10, 15};
    int n = sizeof(arr) / sizeof (arr[0]);
    cout << pairAndSum(arr, n) << endl;
    return 0;
}

Output:

15

This article is contributed by Ekta Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



3.6 Average Difficulty : 3.6/5.0
Based on 27 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.