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

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  • Difficulty Level : Medium
  • Last Updated : 25 Nov, 2021

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

 

Approach: Iterate through all the possible pairs (i, j) and check the count of set bits in their AND 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 AND value is B[j]
            if (__builtin_popcount(A[i] & A[j]) == B[j]) {
                cnt++;
            }
 
    return cnt;
}
 
// Driver code
int main()
{
    int A[] = { 2, 3, 1, 4, 5 };
    int B[] = { 2, 2, 1, 4, 2 };
    int size = sizeof(A) / sizeof(A[0]);
 
    cout << solve(A, B, size);
 
    return 0;
}

Java




// Java implementation of the approach
public 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 AND 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[] = {2, 3, 1, 4, 5};
        int B[] = {2, 2, 1, 4, 2};
        int size = A.length;
 
        System.out.println(solve(A, B, size));
    }
}
 
/* This code contributed by PrinciRaj1992 */

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 AND value is B[j]
            c = A[i] & A[j]
            if (bin(c).count('1') == B[j]) :
                cnt += 1;
    return cnt;
 
# Driver code
if __name__ == "__main__" :
 
    A = [ 2, 3, 1, 4, 5 ];
    B = [ 2, 2, 1, 4, 2 ];
     
    size = len(A);
 
    print(solve(A, B, size));
 
# This code is contributed
# by AnkitRai01

C#




// C# Implementation of the above 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 AND value is B[j]
            {
                if (countSetBits(A[i] & A[j]) == B[j])
                {
                    cnt++;
                }
            }
        }
 
        return cnt;
    }
     
    // Function to get no of set
    // bits in binary representation
    // of positive integer n
    static int countSetBits(int n)
    {
        int count = 0;
        while (n > 0)
        {
            count += n & 1;
            n >>= 1;
        }
        return count;
    }
     
    // Driver code
    public static void Main(String[] args)
    {
        int []A = {2, 3, 1, 4, 5};
        int []B = {2, 2, 1, 4, 2};
        int size = A.Length;
 
        Console.WriteLine(solve(A, B, size));
    }
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript Implementation of the above approach
 
// Function to return the count of pairs
// which satisfy the given condition
function solve(A, B, n)
{
    var cnt = 0;
    for (var i = 0; i < n; i++)
    {
        for (var j = i; j < n; j++)
         
        // Check if the count of set bits
        // in the AND value is B[j]
        {
            if (countSetBits(A[i] & A[j]) == B[j])
            {
                cnt++;
            }
        }
    }
    return cnt;
}
 
// Function to get no of set
// bits in binary representation
// of positive integer n
function countSetBits(n)
{
    var count = 0;
    while (n > 0)
    {
        count += n & 1;
        n >>= 1;
    }
    return count;
}
 
// Driver code
var A = [2, 3, 1, 4, 5];
var B = [2, 2, 1, 4, 2];
var size = A.length;
document.write(solve(A, B, size));
 
// This code is contributed by rutvik_56.
</script>

Output: 

4

 

Time Complexity: O(n2)

Auxiliary Space: O(1)


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