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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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 */

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Output:

7


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