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] = 8Input: 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:
- Traverse the given array.
- For every array element, check if it is a power of 2 or not. Increase the count of such elements
- Finally, print (count * (count – 1)) / 2 as the required count.
Below is the implementation of the above approach:
// 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 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 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# 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 |
<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> |
3
Time Complexity: O(N)
Auxiliary Space: O(1)