Given an array **arr[]** of **N** positive integers. The task is to find the number of pairs whose Bitwise AND value is a power of 2.**Examples:**

Input:arr[] = {2, 1, 3, 4}Output:2Explanation:

There are 2 pairs (2, 3) and (1, 3) in this array whose Bitwise AND values are:

1. (2 & 3) = 1 = (2^{0})

2. (1 & 3) = 1 = (2^{0}).Input:arr[] = {6, 4, 2, 3}Output:4Explanation:

There are 4 pairs (6, 4), (6, 2), (6, 3), (2, 3) whose Bitwise and is power of 2.

**Approach:** For each possible pair in the given array, the idea to check whether **Bitwise AND** of each pairs of elements is **perfect power of 2** or not. If **“Yes”** then count this pair Else check for the next pair.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if x is power of 2` `bool` `check(` `int` `x)` `{` ` ` `// Returns true if x is a power of 2` ` ` `return` `x && (!(x & (x - 1)));` `}` `// Function to return the` `// number of valid pairs` `int` `count(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `cnt = 0;` ` ` `// Iterate for all possible pairs` ` ` `for` `(` `int` `i = 0; i < n - 1; i++) {` ` ` `for` `(` `int` `j = i + 1; j < n; j++) {` ` ` `// Bitwise and value of` ` ` `// the pair is passed` ` ` `if` `(check(arr[i]` ` ` `& arr[j]))` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` `// Return the final count` ` ` `return` `cnt;` `}` `// Driver Code` `int` `main()` `{` ` ` `// Given array` ` ` `int` `arr[] = { 6, 4, 2, 3 };` ` ` `int` `n = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function Call` ` ` `cout << count(arr, n);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `class` `GFG{` `// Method to check if x is power of 2` `static` `boolean` `check(` `int` `x)` `{` ` ` `// First x in the below expression` ` ` `// is for the case when x is 0` ` ` `return` `x != ` `0` `&& ((x & (x - ` `1` `)) == ` `0` `);` `}` `// Function to return the` `// number of valid pairs` `static` `int` `count(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `cnt = ` `0` `;` ` ` `// Iterate for all possible pairs` ` ` `for` `(` `int` `i = ` `0` `; i < n - ` `1` `; i++)` ` ` `{` ` ` `for` `(` `int` `j = i + ` `1` `; j < n; j++)` ` ` `{` ` ` ` ` `// Bitwise and value of` ` ` `// the pair is passed` ` ` `if` `(check(arr[i] & arr[j]))` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` ` ` `// Return the final count` ` ` `return` `cnt;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `// Given array arr[]` ` ` `int` `arr[] = ` `new` `int` `[]{ ` `6` `, ` `4` `, ` `2` `, ` `3` `};` ` ` `int` `n = arr.length;` ` ` ` ` `// Function call` ` ` `System.out.print(count(arr, n));` `}` `}` `// This code is contributed by Pratima Pandey` |

## Python3

`# Python3 program for the above approach` `# Function to check if x is power of 2` `def` `check(x):` ` ` ` ` `# Returns true if x is a power of 2` ` ` `return` `x ` `and` `(` `not` `(x & (x ` `-` `1` `)))` `# Function to return the` `# number of valid pairs` `def` `count(arr, n):` ` ` ` ` `cnt ` `=` `0` ` ` `# Iterate for all possible pairs` ` ` `for` `i ` `in` `range` `(n ` `-` `1` `):` ` ` `for` `j ` `in` `range` `(i ` `+` `1` `, n):` ` ` `# Bitwise and value of` ` ` `# the pair is passed` ` ` `if` `check(arr[i] & arr[j]):` ` ` `cnt ` `=` `cnt ` `+` `1` ` ` `# Return the final count` ` ` `return` `cnt` `# Given array` `arr ` `=` `[ ` `6` `, ` `4` `, ` `2` `, ` `3` `]` `n ` `=` `len` `(arr)` `# Function Call` `print` `(count(arr, n))` `# This code is contributed by divyeshrabadiya07` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Method to check if x is power of 2` `static` `bool` `check(` `int` `x)` `{` ` ` ` ` `// First x in the below expression` ` ` `// is for the case when x is 0` ` ` `return` `x != 0 && ((x & (x - 1)) == 0);` `}` `// Function to return the` `// number of valid pairs` `static` `int` `count(` `int` `[]arr, ` `int` `n)` `{` ` ` `int` `cnt = 0;` ` ` `// Iterate for all possible pairs` ` ` `for` `(` `int` `i = 0; i < n - 1; i++)` ` ` `{` ` ` `for` `(` `int` `j = i + 1; j < n; j++)` ` ` `{` ` ` ` ` `// Bitwise and value of` ` ` `// the pair is passed` ` ` `if` `(check(arr[i] & arr[j]))` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` ` ` `// Return the final count` ` ` `return` `cnt;` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` ` ` `// Given array arr[]` ` ` `int` `[]arr = ` `new` `int` `[]{ 6, 4, 2, 3 };` ` ` `int` `n = arr.Length;` ` ` ` ` `// Function call` ` ` `Console.Write(count(arr, n));` `}` `}` `// This code is contributed by Code_Mech` |

## Javascript

`<script>` `// JavaScript program to implement` `// the above approach` `// Method to check if x is power of 2` `function` `check(x)` `{` ` ` ` ` `// First x in the below expression` ` ` `// is for the case when x is 0` ` ` `return` `x != 0 && ((x & (x - 1)) == 0);` `}` ` ` `// Function to return the` `// number of valid pairs` `function` `count(arr, n)` `{` ` ` `let cnt = 0;` ` ` ` ` `// Iterate for all possible pairs` ` ` `for` `(let i = 0; i < n - 1; i++)` ` ` `{` ` ` `for` `(let j = i + 1; j < n; j++)` ` ` `{` ` ` ` ` `// Bitwise and value of` ` ` `// the pair is passed` ` ` `if` `(check(arr[i] & arr[j]))` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` ` ` `// Return the final count` ` ` `return` `cnt;` `}` `// Driver code` ` ` `// Given array arr[]` ` ` `let arr = [ 6, 4, 2, 3 ];` ` ` ` ` `let n = arr.length;` ` ` ` ` `// Function call` ` ` `document.write(count(arr, n));` `// This code is contributed by susmitakundugoaldanga.` `</script>` |

**Output:**

4

**Time Complexity:** O(N^{2}) **Auxiliary Space:** O(1)

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