Java Program for Pairs such that one is a power multiple of other

We 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:

// Sorting 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++;
        }
    }
}
// Returning 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;
}

// Main 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