Skip to content
Related Articles

Related Articles

Improve Article

Number of pairs whose product is a power of 2

  • Last Updated : 22 Jun, 2021

Given an array arr[] consisting of N integers, the task is to count the total number of pairs of array elements from the given array such that arr[i] * arr[j] is the power of 2.

Examples:

Input: arr[] = {2, 4, 7, 2}
Output: 3
Explanation:
arr[0] * arr[1] = 8
arr[0] * arr[3] = 4
arr[1] * arr[3] = 8

Input: arr[] = {8, 1, 12, 4, 2}
Output: 6

 

Approach: The idea is based upon the fact that a number is a power of 2 if it contains only 2 as its prime factor. Therefore, all its divisors are also a power of 2. Follow the steps below to solve the problem:



  1. Traverse the given array.
  2. For every array element, check if it is a power of 2 or not. Increase the count of such elements
  3. Finally, print (count * (count – 1)) / 2 as the required count.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count pairs having
// product equal to a power of 2
int countPairs(int arr[], int N)
{
    // Stores count of array elements
    // which are power of 2
    int countPowerof2 = 0;
 
    for (int i = 0; i < N; i++) {
 
        // If array element contains
        // only one set bit
        if (__builtin_popcount(arr[i]) == 1)
 
            // Increase count of
            // powers of 2
            countPowerof2++;
    }
 
    // Count required number of pairs
    int desiredPairs
        = (countPowerof2
           * (countPowerof2 - 1))
          / 2;
 
    // Print the required number of pairs
    cout << desiredPairs << ' ';
}
 
// Driver Code
int main()
{
    // Given array
    int arr[4] = { 2, 4, 7, 2 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    countPairs(arr, N);
 
    return 0;
}

Java




// Java program for the
// above approach
import java.util.*;
class GFG{
 
// Function to count pairs having
// product equal to a power of 2
static void countPairs(int arr[],
                       int N)
{
  // Stores count of array elements
  // which are power of 2
  int countPowerof2 = 0;
 
  for (int i = 0; i < N; i++)
  {
    // If array element contains
    // only one set bit
    if (Integer.bitCount(arr[i]) == 1)
 
      // Increase count of
      // powers of 2
      countPowerof2++;
  }
 
  // Count required number of pairs
  int desiredPairs = (countPowerof2 *
                     (countPowerof2 - 1)) / 2;
 
  // Print the required number of pairs
  System.out.print(desiredPairs + " ");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given array
  int arr[] = {2, 4, 7, 2};
 
  // Size of the array
  int N = arr.length;
 
  // Function Call
  countPairs(arr, N);
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program for the above approach
 
# Function to count pairs having
# product equal to a power of 2
def countPairs(arr, N):
     
    # Stores count of array elements
    # which are power of 2
    countPowerof2 = 0
 
    for i in range(N):
 
        # If array element contains
        # only one set bit
        if (bin(arr[i]).count('1') == 1):
 
            # Increase count of
            # powers of 2
            countPowerof2 += 1
 
    # Count required number of pairs
    desiredPairs = (countPowerof2 *
                   (countPowerof2 - 1)) // 2
 
    # Print the required number of pairs
    print(desiredPairs)
 
# Driver Code
if __name__ == '__main__':
     
    # Given array
    arr = [ 2, 4, 7, 2 ]
 
    # Size of the array
    N = len(arr)
 
    # Function call
    countPairs(arr, N)
 
# This code is contributed by mohit kumar 29

C#




// C# program for the
// above approach
using System;
using System.Linq;
 
class GFG{
 
// Function to count pairs having
// product equal to a power of 2
static void countPairs(int []arr,
                       int N)
{
     
    // Stores count of array elements
    // which are power of 2
    int countPowerof2 = 0;
     
    for(int i = 0; i < N; i++)
    {
         
        // If array element contains
        // only one set bit
        if ((Convert.ToString(
             arr[i], 2)).Count(
             f => (f == '1')) == 1)
              
            // Increase count of
            // powers of 2
            countPowerof2++;
    }
     
    // Count required number of pairs
    int desiredPairs = (countPowerof2 *
                       (countPowerof2 - 1)) / 2;
     
    // Print the required number of pairs
    Console.WriteLine(desiredPairs + " ");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array
    int []arr = { 2, 4, 7, 2 };
     
    // Size of the array
    int N = arr.Length;
     
    // Function call
    countPairs(arr, N);
}
}
 
// This code is contributed by math_lover

Javascript




<script>
// Javascript program for the
// above approach
 
// Function to count pairs having
// product equal to a power of 2
function countPairs(arr,N)
{
    // Stores count of array elements
  // which are power of 2
  let countPowerof2 = 0;
  
  for (let i = 0; i < N; i++)
  {
    // If array element contains
    // only one set bit
    if (Number(arr[i].toString(2).split("").sort().join("")).toString().length == 1)
  
      // Increase count of
      // powers of 2
      countPowerof2++;
  }
  
  // Count required number of pairs
  let desiredPairs = (countPowerof2 *
                     (countPowerof2 - 1)) / 2;
  
  // Print the required number of pairs
  document.write(desiredPairs + " ");
}
 
// Driver Code
// Given array
let arr=[2, 4, 7, 2];
// Size of the array
let N = arr.length;
 
// Function Call
countPairs(arr, N);
 
 
// This code is contributed by patel2127
</script>
Output: 
3

 

Time Complexity: O(N)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :