# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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); ` ` `  `    ``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 */`

Output:

```7
```

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