# Count pairs of elements such that number of set bits in their AND 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[] = {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 ``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

 ``

Output:
`4`

Time Complexity: O(n2)

Auxiliary Space: O(1)

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