Given an array of N elements. The task is to count unordered pairs (i, j) in the array such that the product of a[i] and a[j] can be expressed as a power of two.
Examples:
Input : arr[] = {2, 3, 4, 8, 10}
Output : 3
Explanation: The pair of array element will be
(2, 4), (2, 8), (4, 8) whose product are
8, 16, 32 respectively which can be expressed
as power of 2, like 2^3, 2^4, 2^5.
Input : arr[] = { 2, 5, 8, 16, 128 }
Output : 6
If you multiply 
and and their product become , then z=x*y, now if it’s possible to express as power of two then it can be proved that both and can be expressed as power of two. Basically z= 2a = 2(b+c) = 2b * 2c = x * y, where and both can hold a minimum value 0.
So now we have to count the number of elements in the array which can be expressed as a power of two. If the count is k, then answer will be kC2 = k*(k-1)/2, as we need the count of unordered pairs.
Below is the implementation of above approach:
C++
// C++ program to Count unordered pairs (i, j)// in array such that product of a[i] and a[j]// can be expressed as power of two#include <bits/stdc++.h>using namespace std;/* Function to check if x is power of 2*/bool isPowerOfTwo(int x) { /* First x in the below expression is for the case when x is 0 */ return x && (!(x&(x-1))); } // Function to Count unordered pairsvoid Count_pairs(int a[], int n){ int count = 0; for (int i = 0; i < n; i++) { // is a number can be expressed // as power of two if (isPowerOfTwo(a[i])) count++; } // count total number // of unordered pairs int ans = (count * (count - 1)) / 2; cout << ans << "\n";}// Driver codeint main(){ int a[] = { 2, 5, 8, 16, 128 }; int n = sizeof(a) / sizeof(a[0]); Count_pairs(a, n); return 0;} |
Java
// Java program to Count unordered pairs (i, j)// in array such that product of a[i] and a[j]// can be expressed as power of twoimport java.io.*;class GFG {/* Function to check if x is power of 2*/static boolean isPowerOfTwo(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 Count unordered pairsstatic void Count_pairs(int a[], int n){ int count = 0; for (int i = 0; i < n; i++) { // is a number can be expressed // as power of two if (isPowerOfTwo(a[i])) count++; } // count total number // of unordered pairs int ans = (count * (count - 1)) / 2; System.out.println( ans);}// Driver code public static void main (String[] args) { int a[] = { 2, 5, 8, 16, 128 }; int n = a.length; Count_pairs(a, n); }}// This code is contributed// by shs |
Python 3
# Python3 program to Count unordered pairs # (i, j) in array such that product of a[i] # and a[j] can be expressed as power of two # Function to check if x is power of 2def isPowerOfTwo(x) : # First x in the below expression # is for the case when x is 0 return (x and(not(x & (x - 1))))# Function to Count unordered pairs def Count_pairs(a, n) : count = 0 for i in range(n) : # is a number can be expressed # as power of two if isPowerOfTwo(a[i]) : count += 1 # count total number # of unordered pairs ans = (count * (count - 1)) / 2 print(ans)# Driver code if __name__ == "__main__" : a = [ 2, 5, 8, 16, 128] n = len(a) Count_pairs(a, n) # This code is contributed by ANKITRAI1 |
C#
// C# program to Count unordered pairs (i, j) // in array such that product of a[i] and a[j] // can be expressed as power of two using System;public class GFG{ /* Function to check if x is power of 2*/static bool isPowerOfTwo(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 Count unordered pairs static void Count_pairs(int []a, int n) { int count = 0; for (int i = 0; i < n; i++) { // is a number can be expressed // as power of two if (isPowerOfTwo(a[i])) count++; } // count total number // of unordered pairs int ans = (count * (count - 1)) / 2; Console.WriteLine( ans); } // Driver code static public void Main (){ int []a = { 2, 5, 8, 16, 128 }; int n = a.Length; Count_pairs(a, n); } } // This code is contributed // by Sach_Code |
PHP
<?php// PHP program to Count unordered // pairs (i, j) in array such that // product of a[i] and a[j] can be // expressed as power of two /* Function to check if x is power of 2*/function isPowerOfTwo($x) { /* First x in the below expression is for the case when x is 0 */ return ($x && (!($x & ($x - 1)))); } // Function to Count unordered pairs function Count_pairs($a, $n) { $count = 0; for ($i = 0; $i < $n; $i++) { // is a number can be expressed // as power of two if (isPowerOfTwo($a[$i])) $count++; } // count total number // of unordered pairs $ans = ($count * ($count - 1)) / 2; echo $ans , "\n"; } // Driver code $a = array( 2, 5, 8, 16, 128 ); $n = sizeof($a); Count_pairs($a, $n); // This code is contributed // by Sach_code?> |
Javascript
<script>// JavaScript program to // Count unordered pairs (i, j)// in array such that product of a[i] and a[j]// can be expressed as power of two/* Function to check if x is power of 2*/function isPowerOfTwo( x){/* First x in the below expression is for the case when x is 0 */return (x >0&& (!((x&(x-1))>0)));}// Function to Count unordered pairsfunction Count_pairs(a,n){ let count = 0; for (let i = 0; i < n; i++) { // is a number can be expressed // as power of two if (isPowerOfTwo(a[i])) count++; } // count total number // of unordered pairs let ans = (count * (count - 1)) / 2; document.write( ans);}// Driver code let a = [ 2, 5, 8, 16, 128 ]; let n = a.length; Count_pairs(a, n);// This code is contributed by sravan kumar</script> |
Output
6
Complexity Analysis:
- Time Complexity: O(N), where N is the number of elements in the array.
- Auxiliary Space: O(1)
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