Count pairs of elements such that number of set bits in their OR is B[i]

Given two arrays A[] and B[] of N elements each. The task is to find the number of index pairs (i, j) such that i ? j and F(A[i] | A[j]) = B[j] where F(X) is the count of set bits in the binary representation of X.
Examples

Input: A[] = {5, 3, 2, 4, 6, 1}, B[] = {2, 2, 1, 4, 2, 3} 
Output: 7 
All possible pairs are (5, 5), (3, 3), (2, 2), 
(2, 6), (4, 6), (6, 6) and (6, 1).
Input: A[] = {4, 3, 5, 6, 7}, B[] = {1, 3, 2, 4, 5} 
Output: 4 

Approach: Iterate through all the possible pairs (i, j) and check the count of set bits in their OR value. If the count is equal to B[j] then increment the count.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the count of pairs
// which satisfy the given condition
int solve(int A[], int B[], int n)
{
    int cnt = 0;

    for (int i = 0; i < n; i++)
        for (int j = i; j < n; j++)

            // Check if the count of set bits
            // in the OR value is B[j]
            if (__builtin_popcount(A[i] | A[j]) == B[j]) {
                cnt++;
            }

    return cnt;
}

// Driver code
int main()
{
    int A[] = { 5, 3, 2, 4, 6, 1 };
    int B[] = { 2, 2, 1, 4, 2, 3 };
    int size = sizeof(A) / sizeof(A[0]);

    cout << solve(A, B, size);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the count of pairs
// which satisfy the given condition
static int solve(int A[], int B[], int n)
{
    int cnt = 0;

    for (int i = 0; i < n; i++)
        for (int j = i; j < n; j++)

            // Check if the count of set bits
            // in the OR value is B[j]
            if (Integer.bitCount(A[i] | A[j]) == B[j])
            {
                cnt++;
            }

    return cnt;
}

// Driver code
public static void main(String args[])
{
    int A[] = { 5, 3, 2, 4, 6, 1 };
    int B[] = { 2, 2, 1, 4, 2, 3 };
    int size = A.length;

    System.out.println(solve(A, B, size));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Function to return the count of pairs
# which satisfy the given condition
def solve(A, B, n) :

    cnt = 0;
    for i in range(n) :
        for j in range(i, n) :

            # Check if the count of set bits
            # in the OR value is B[j]
            if (bin(A[i] | A[j]).count('1') == B[j]) :
                cnt += 1;

    return cnt


# Driver code
if __name__ == "__main__" :

    A = [ 5, 3, 2, 4, 6, 1 ];
    B = [ 2, 2, 1, 4, 2, 3 ];
    size = len(A);

    print(solve(A, B, size));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the count of pairs
// which satisfy the given condition
static int solve(int []A, int []B, int n)
{
    int cnt = 0;

    for (int i = 0; i < n; i++)
        for (int j = i; j < n; j++)

            // Check if the count of set bits
            // in the OR value is B[j]
            if (bitCount(A[i] | A[j]) == B[j])
            {
                cnt++;
            }

    return cnt;
}

static int bitCount(long x)
{
    // To store the count
    // of set bits
    int setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}

// Driver code
public static void Main(String []args)
{
    int []A = { 5, 3, 2, 4, 6, 1 };
    int []B = { 2, 2, 1, 4, 2, 3 };
    int size = A.Length;

    Console.WriteLine(solve(A, B, size));
}
}

/* This code is contributed by PrinciRaj1992 */

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the count of pairs
// which satisfy the given condition
function solve(A,B,n)
{
    let cnt = 0;

    for (let i = 0; i < n; i++)
        for (let j = i; j < n; j++)

            // Check if the count of set bits
            // in the OR value is B[j]
            if (bitCount(A[i] | A[j]) == B[j])
            {
                cnt++;
            }

    return cnt;
}

function bitCount(x)
{
    // To store the count
    // of set bits
    let setBits = 0;
    while (x != 0)
    {
        x = x & (x - 1);
        setBits++;
    }
    return setBits;
}


// Driver code
let A=[5, 3, 2, 4, 6, 1 ];
let B=[2, 2, 1, 4, 2, 3 ];
let size = A.length;
document.write(solve(A, B, size));


// This code is contributed by rag2127

</script>

Output:

7

Time Complexity: O(N²)

Auxiliary Space: O(1)