# Sum of Bitwise And of all pairs in a given array

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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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.
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`

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