Count of pairs whose bitwise AND is a power of 2

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: 2 
Explanation:
There are 2 pairs (2, 3) and (1, 3) in this array whose Bitwise AND values are: 
1. (2 & 3) = 1 = (2⁰) 
2. (1 & 3) = 1 = (2⁰).
Input: arr[] = {6, 4, 2, 3} 
Output: 4 
Explanation:
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(N²) 
Auxiliary Space: O(1)