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

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

C++

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// 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;
}

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Java

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// A Simple Java program to compute
// sum of bitwise AND of all pairs
import java.io.*;
  
class GFG {
      
    // Returns value of "arr[0] & arr[1] +
    // arr[0] & arr[2] + ... arr[i] & arr[j] + 
    // ..... arr[n-2] & arr[n-1]"
    static 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
    public static void main(String args[])
    {
        int arr[] = {5, 10, 15};
        int n = arr.length;
        System.out.println(pairAndSum(arr, n) );
    }
}
  
/*This code is contributed by Nikita Tiwari.*/

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Python3

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# A Simple Python 3 program to compute
# sum of bitwise AND of all pairs
  
# Returns value of "arr[0] & arr[1] +
# arr[0] & arr[2] + ... arr[i] & arr[j] +
# ..... arr[n-2] & arr[n-1]"
def pairAndSum(arr, n) :
    ans = 0 # Initialize result
  
    # Consider all pairs (arr[i], arr[j) 
    # such that i < j
    for i in range(0,n) :
        for j in range((i+1),n) :
            ans = ans + arr[i] & arr[j]
  
    return ans
  
# Driver program to test above function
arr = [5, 10, 15]
n = len(arr) 
print(pairAndSum(arr, n))
  
# This code is contributed by Nikita Tiwari.

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C#

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// A Simple C# program to compute
// sum of bitwise AND of all pairs
using System;
  
class GFG {
       
    // Returns value of "arr[0] & arr[1] +
    // arr[0] & arr[2] + ... arr[i] & arr[j] + 
    // ..... arr[n-2] & arr[n-1]"
    static 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
    public static void Main()
    {
        int []arr = {5, 10, 15};
        int n = arr.Length;
        Console.Write(pairAndSum(arr, n) );
    }
}
   
// This code is contributed by nitin mittal.

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PHP

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<?php
// A Simple PHP program to 
// compute sum of bitwise 
// AND of all pairs
  
// Returns value of "arr[0] &
// arr[1] + arr[0] & arr[2] + 
// ... arr[i] & arr[j] + .....
// arr[n-2] & arr[n-1]"
  
function pairAndSum($arr, $n)
{
    // Initialize result
    $ans = 0; 
  
    // Consider all pairs (arr[i], 
    // arr[j) such that i < j
    for ($i = 0; $i < $n; $i++)
        for ( $j = $i + 1; $j < $n; $j++)
        $ans += $arr[$i] & $arr[$j];
  
    return $ans;
}
  
// Driver Code
$arr = array(5, 10, 15);
$n = sizeof($arr) ;
echo pairAndSum($arr, $n), "\n";
  
// This code is contributed by m_kit
?>

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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 the implementation.

C

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// 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;
}

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Java

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// An efficient Java program to compute
// sum of bitwise AND of all pairs
import java.io.*;
  
class GFG {
      
    // Returns value of "arr[0] & arr[1] + 
    // arr[0] & arr[2] + ... arr[i] & arr[j] +
    // ..... arr[n-2] & arr[n-1]"
    static 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
            // Initialize the count
            int k = 0;
            for (int j = 0; j < n; j++)
            {
                if ((arr[j] & (1 << i))!=0)
                    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
    public static void main(String args[])
    {
        int arr[] = {5, 10, 15};
        int n = arr.length;
        System.out.println(pairAndSum(arr, n));
    }
}
  
/*This code is contributed by Nikita Tiwari.*/

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Python3

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# An efficient Python 3 program to
# compute sum of bitwise AND of all pairs
  
# Returns value of "arr[0] & arr[1] +
# arr[0] & arr[2] + ... arr[i] & arr[j] +
# ..... arr[n-2] & arr[n-1]"
def pairAndSum(arr, n) :
    ans = 0 # Initialize result
  
    # Traverse over all bits
    for i in range(0,32) :
          
        # Count number of elements with i'th bit set
        # Initialize the count
        k = 0
        for j in range(0,n) :
            if ( (arr[j] & (1 << i)) ) :
                k = k + 1
  
        # 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 = ans + (1 << i) * (k * (k - 1) // 2)
      
    return ans
      
# Driver program to test above function
arr = [5, 10, 15]
n = len(arr) 
print(pairAndSum(arr, n))
  
# This code is contributed by Nikita Tiwari.

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C#

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// An efficient C# program to compute
// sum of bitwise AND of all pairs
using System;
  
class GFG {
      
    // Returns value of "arr[0] & arr[1] + 
    // arr[0] & arr[2] + ... arr[i] & arr[j] +
    // ..... arr[n-2] & arr[n-1]"
    static 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 Initialize the count
            int k = 0;
            for (int j = 0; j < n; j++)
            {
                if ((arr[j] & (1 << i))!=0)
                    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
    public static void Main()
    {
        int []arr = new int[]{5, 10, 15};
        int n = arr.Length;
          
        Console.Write(pairAndSum(arr, n));
    }
}
  
/* This code is contributed by smitha*/

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PHP

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<?php
// An efficient PHP program to 
// compute sum of bitwise AND
// of all pairs
  
// Returns value of "arr[0] & 
// arr[1] + arr[0] & arr[2] + 
// ... arr[i] & arr[j] + ..... 
// arr[n-2] & arr[n-1]"
function pairAndSum($arr, $n)
{
    // Initialize result
    $ans = 0; 
  
    // Traverse over all bits
    for ($i = 0; $i < 32; $i++)
    {
          
        // Count number of elements
        // with i'th bit set
        // Initialize the count
        $k = 0; 
        for ($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 Code
    $arr = array(5, 10, 15);
    $n = sizeof($arr);
    echo pairAndSum($arr, $n) ;
  
// This code is contributed by nitin mittal.
?>

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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.



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