Count unordered pairs (i,j) such that product of a[i] and a[j] is power of two

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:



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// 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 pairs
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;
  
    cout << ans << "\n";
}
  
// Driver code
int main()
{
    int a[] = { 2, 5, 8, 16, 128 };
  
    int n = sizeof(a) / sizeof(a[0]);
  
    Count_pairs(a, n);
  
    return 0;
}
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// 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 two
  
import 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 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;
  
    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
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# 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 2
def 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
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// 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
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<?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
?>
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Output:
6

Time Complexity: O(N), where N is the number of elements in the array.

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Improved By : Shashank12, Sach_Code, AnkitRai01

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