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Pairs such that one is a power multiple of other

• Difficulty Level : Easy
• Last Updated : 30 Apr, 2021

You are given an array A[] of n-elements and a positive integer k. Now you have find the number of pairs Ai, Aj such that Ai = Aj*(kx) where x is an integer.
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 ans;```

C++

 `// Program to find pairs count``#include ``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

 `// 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.`

Python3

 `# 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#

 `// 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

 ``

Javascript

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Output :
`6`

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