Pairs such that one is a power multiple of other

You are given an array A[] of n-elements and a positive integer k(other than 1). Now you have find the number of pairs Ai, Aj such that Ai = Aj*(k^x) where x is an integer. Given that (k?1).

Note: (Ai, Aj) and (Aj, Ai) must be count once.

Examples : 

Input : A[] = {3, 6, 4, 2},  k = 2
Output : 2
Explanation : We have only two pairs 
(4, 2) and (3, 6)

Input : A[] = {2, 2, 2},   k = 2
Output : 3
Explanation : (2, 2), (2, 2), (2, 2) 
that are (A1, A2), (A2, A3) and (A1, A3) are 
total three pairs where Ai = Aj * (k^0) 

To solve this problem, we first sort the given array and then for each element Ai, we find number of elements equal to value Ai * k^x for different value of x till Ai * k^x is less than or equal to largest of Ai. 

Algorithm: 

    // sort the given array
    sort(A, A+n);

    // for each A[i] traverse rest array
    for (int i=0; i<n; i++)
    {
        for (int j=i+1; j<n; j++)
        {
            // count Aj such that Ai*k^x = Aj
            int x = 0;

            // increase x till Ai * k^x <= 
            // largest element
            while ((A[i]*pow(k, x)) <= A[j])
            {
                if ((A[i]*pow(k, x)) == A[j])
                {              
                     ans++;
                     break;
                }
                x++;
            }        
        }   
    }
    // return answer
    return ans;

Implementation:

// Program to find pairs count
#include <bits/stdc++.h>
using namespace std;

// function to count the required pairs
int countPairs(int A[], int n, int k) {
  int ans = 0;
  // sort the given array
  sort(A, A + n);

  // for each A[i] traverse rest array
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {

      // count Aj such that Ai*k^x = Aj
      int x = 0;

      // increase x till Ai * k^x <= largest element
      while ((A[i] * pow(k, x)) <= A[j]) {
        if ((A[i] * pow(k, x)) == A[j]) {
          ans++;
          break;
        }
        x++;
      }
    }
  }
  return ans;
}

// driver program
int main() {
  int A[] = {3, 8, 9, 12, 18, 4, 24, 2, 6};
  int n = sizeof(A) / sizeof(A[0]);
  int k = 3;
  cout << countPairs(A, n, k);
  return 0;
}

// Java program to find pairs count
import java.io.*;
import java .util.*;

class GFG {
    
    // function to count the required pairs
    static int countPairs(int A[], int n, int k) 
    {
        int ans = 0;
        
        // sort the given array
        Arrays.sort(A);
        
        // for each A[i] traverse rest array
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) 
            {
        
                // count Aj such that Ai*k^x = Aj
                int x = 0;
            
                // increase x till Ai * k^x <= largest element
                while ((A[i] * Math.pow(k, x)) <= A[j]) 
                {
                    if ((A[i] * Math.pow(k, x)) == A[j]) 
                    {
                        ans++;
                        break;
                    }
                    x++;
                }
            }
        }
        return ans;
    }
    
    // Driver program
    public static void main (String[] args) 
    {
        int A[] = {3, 8, 9, 12, 18, 4, 24, 2, 6};
        int n = A.length;
        int k = 3;
        System.out.println (countPairs(A, n, k));
        
    }
}

// This code is contributed by vt_m.

# Program to find pairs count
import math

# function to count the required pairs
def countPairs(A, n, k): 
    ans = 0

    # sort the given array
    A.sort()
    
    # for each A[i] traverse rest array
    for i in range(0,n): 

        for j in range(i + 1, n):

            # count Aj such that Ai*k^x = Aj
            x = 0

            # increase x till Ai * k^x <= largest element
            while ((A[i] * math.pow(k, x)) <= A[j]) :
                if ((A[i] * math.pow(k, x)) == A[j]) :
                    ans+=1
                    break
                x+=1
    return ans


# driver program
A = [3, 8, 9, 12, 18, 4, 24, 2, 6]
n = len(A)
k = 3

print(countPairs(A, n, k))

# This code is contributed by
# Smitha Dinesh Semwal

// C# program to find pairs count
using System;

class GFG {
    
    // function to count the required pairs
    static int countPairs(int []A, int n, int k) 
    {
        int ans = 0;
        
        // sort the given array
        Array.Sort(A);
        
        // for each A[i] traverse rest array
        for (int i = 0; i < n; i++) 
        {
            for (int j = i + 1; j < n; j++) 
            {
        
                // count Aj such that Ai*k^x = Aj
                int x = 0;
            
                // increase x till Ai * k^x <= largest element
                while ((A[i] * Math.Pow(k, x)) <= A[j]) 
                {
                    if ((A[i] * Math.Pow(k, x)) == A[j]) 
                    {
                        ans++;
                        break;
                    }
                    x++;
                }
            }
        }
        return ans;
    }
    
    // Driver program
    public static void Main () 
    {
        int []A = {3, 8, 9, 12, 18, 4, 24, 2, 6};
        int n = A.Length;
        int k = 3;
        Console.WriteLine(countPairs(A, n, k));
        
    }
}

// This code is contributed by vt_m.

<?php
// PHP Program to find pairs count

// function to count
// the required pairs
function countPairs($A, $n, $k) 
{
$ans = 0;

// sort the given array
sort($A);

// for each A[i] 
// traverse rest array
for ($i = 0; $i < $n; $i++) 
{
    for ($j = $i + 1; $j < $n; $j++) 
    {

    // count Aj such that Ai*k^x = Aj
    $x = 0;

    // increase x till Ai * 
    // k^x <= largest element
    while (($A[$i] * pow($k, $x)) <= $A[$j]) 
    {
        if (($A[$i] * pow($k, $x)) == $A[$j]) 
        {
        $ans++;
        break;
        }
        $x++;
    }
    }
}
return $ans;
}

// Driver Code

$A = array(3, 8, 9, 12, 18, 
              4, 24, 2, 6);
$n = count($A);
$k = 3;
echo countPairs($A, $n, $k);

// This code is contributed by anuj_67.
?>

<script>

// Javascript Program to find pairs count

// function to count the required pairs
function countPairs(A, n, k) {
  var ans = 0;
  
  // sort the given array
  A.sort((a,b)=>a-b)

  // for each A[i] traverse rest array
  for (var i = 0; i < n; i++) {
    for (var j = i + 1; j < n; j++) {

      // count Aj such that Ai*k^x = Aj
      var x = 0;

      // increase x till Ai * k^x <= largest element
      while ((A[i] * Math.pow(k, x)) <= A[j]) {
        if ((A[i] * Math.pow(k, x)) == A[j]) {
          ans++;
          break;
        }
        x++;
      }
    }
  }
  return ans;
}

// driver program
var A = [3, 8, 9, 12, 18, 4, 24, 2, 6];
var n = A.length;
var k = 3;
document.write( countPairs(A, n, k));

// This code is contributed by rutvik_56.
</script>  

6

Time Complexity: O(n*n), as nested loops are used
Auxiliary Space: O(1), as no extra space is used