# 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 * arr = 8
arr * arr = 4
arr * arr = 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 ` `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 = { 2, 4, 7, 2 };`   `    ``// Size of the array` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);`   `    ``// 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

 ``

Output:

`3`

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

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