Sum of Bitwise AND of all pairs possible from two arrays
Last Updated :
02 Sep, 2022
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: 3
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++
#include <bits/stdc++.h>
using namespace std;
int sumOfAnd( int A[], int B[],
int N, int M)
{
int pairsAndSum = 0;
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < M;
j++) {
pairsAndSum +=
(A[i] & B[j]);
}
}
return pairsAndSum;
}
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;
}
|
Java
import java.util.*;
class GFG{
static int sumOfAnd( int A[], int B[],
int N, int M)
{
int pairsAndSum = 0 ;
for ( int i = 0 ; i < N; i++)
{
for ( int j = 0 ; j < M; j++)
{
pairsAndSum += (A[i] & B[j]);
}
}
return pairsAndSum;
}
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));
}
}
|
Python3
def sumOfAnd(A, B, N, M):
pairsAndSum = 0
for i in range (N):
for j in range (M):
pairsAndSum + = (A[i] & B[j])
return pairsAndSum
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))
|
C#
using System;
class GFG{
static int sumOfAnd( int []A, int []B,
int N, int M)
{
int pairsAndSum = 0;
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < M; j++)
{
pairsAndSum += (A[i] & B[j]);
}
}
return pairsAndSum;
}
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));
}
}
|
Javascript
<script>
function sumOfAnd(A, B, N, M)
{
var pairsAndSum = 0;
for ( var i = 0; i < N; i++) {
for ( var j = 0; j < M;
j++) {
pairsAndSum +=
(A[i] & B[j]);
}
}
return pairsAndSum;
}
var A = [ 4, 6, 0, 0, 3, 3 ];
var B = [ 0, 5, 6, 5, 0, 3 ];
var N = A.length;
var M = B.length;
document.write(sumOfAnd(A, B, N, M));
</script>
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Time Complexity: O(N2), since two nested loops are used.
Auxiliary Space: O (1), as constant space is being used.
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