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# 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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