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Sum of Bitwise AND of all pairs possible from two arrays
  • Last Updated : 04 Nov, 2020

Given two arrays A[] and B[] of size N and M respectively, the task is to find the sum of Bitwise AND of all possible unordered pairs (A[i], B[j]) from the two arrays.

Examples:

Input: A[] = {1, 2} , B[] = {3, 4} 
Output:
Explanation: 
Bitwise AND of all possible pairs are 
1 & 3  = 1 
1 & 4 = 0 
2 & 3 = 2 
2 & 4 = 0 
Therefore, the sum of bitwise AND of all possible pairs are = (1 + 0 + 2 + 0) = 3

Input: A[] = {4, 6, 0, 0, 3, 3}, B[] = {0, 5, 6, 5, 0, 3} 
Output: 42

 

Approach: To solve the problem, the idea is to traverse both the arrays and generate all possible pairs from the given two arrays and keep adding their respective Bitwise ANDs. Finally, print the sum of Bitwise AND of all possible pairs (A[i], B[j]) obtained from the two given arrays.



Follow the steps below to solve the problem:

  • Initialize a variable, say pairsAndSum to store the sum of Bitwise AND of all possible pairs.
  • Traverse both the array and generate all possible pairs from the given two arrays.
  • Finally, calculate the sum of Bitwise AND of all possible pairs from both the arrays and print the sum.

Below is the implementation of the above approach:

C++

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// C++ Program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the sum of
// AND of all possible pair
int sumOfAnd(int A[], int B[],
                int N, int M)
{
 
    // Stores sum of bitwise AND
    // of  all possible pair
    int pairsAndSum = 0;
 
    // Traverse the array A[]
    for (int i = 0; i < N; i++) {
         
       // Traverse the array B[]
        for (int j = 0; j < M;
                           j++) {
 
            // Update pairsAndSum
            pairsAndSum +=
                   (A[i] & B[j]);
        }
    }
     
    return pairsAndSum;
}
 
// Driver Code
int main()
{
 
    int A[] = { 4, 6, 0, 0, 3, 3 };
    int B[] = { 0, 5, 6, 5, 0, 3 };
 
    int N = sizeof(A) / sizeof(A[0]);
 
    int M = sizeof(B) / sizeof(B[0]);
     
    cout << sumOfAnd(A, B, N, M);
 
    return 0;
}

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Java

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// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to find the sum of
// AND of all possible pair
static int sumOfAnd(int A[], int B[],
                    int N, int M)
{
  // Stores sum of bitwise AND
  // of  all possible pair
  int pairsAndSum = 0;
 
  // Traverse the array A[]
  for (int i = 0; i < N; i++)
  {
    // Traverse the array B[]
    for (int j = 0; j < M; j++)
    {
      // Update pairsAndSum
      pairsAndSum += (A[i] & B[j]);
    }
  }
 
  return pairsAndSum;
}
 
// Driver Code
public static void main(String[] args)
{
  int A[] = {4, 6, 0, 0, 3, 3};
  int B[] = {0, 5, 6, 5, 0, 3};
  int N = A.length;
  int M = B.length;
  System.out.print(sumOfAnd(A, B,
                            N, M));
}
}
 
// This code is contributed by gauravrajput1

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Python3

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# Python3 program to implement
# the above approach
 
# Function to find the sum of
# AND of all possible pair
def sumOfAnd(A, B, N, M):
     
    # Stores sum of bitwise AND
    # of all possible pair
    pairsAndSum = 0
 
    # Traverse the array A
    for i in range(N):
         
        # Traverse the array B
        for j in range(M):
             
            # Update pairsAndSum
            pairsAndSum += (A[i] & B[j])
 
    return pairsAndSum
 
# Driver Code
if __name__ == '__main__':
     
    A = [ 4, 6, 0, 0, 3, 3 ]
    B = [ 0, 5, 6, 5, 0, 3 ]
     
    N = len(A)
    M = len(B)
     
    print(sumOfAnd(A, B, N, M))
 
# This code is contributed by Amit Katiyar

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C#

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// C# Program to implement
// the above approach
using System;
class GFG{
 
// Function to find the sum of
// AND of all possible pair
static int sumOfAnd(int []A, int []B,
                    int N, int M)
{
  // Stores sum of bitwise AND
  // of  all possible pair
  int pairsAndSum = 0;
 
  // Traverse the array []A
  for (int i = 0; i < N; i++)
  {
    // Traverse the array []B
    for (int j = 0; j < M; j++)
    {
      // Update pairsAndSum
      pairsAndSum += (A[i] & B[j]);
    }
  }
 
  return pairsAndSum;
}
 
// Driver Code
public static void Main(String[] args)
{
  int []A = {4, 6, 0, 0, 3, 3};
  int []B = {0, 5, 6, 5, 0, 3};
  int N = A.Length;
  int M = B.Length;
  Console.Write(sumOfAnd(A, B,
                            N, M));
}
}
 
// This code is contributed by Rajput-Ji

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Output: 

42










 

Time Complexity: O(N2)
Auxiliary Space: O(1)

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