Pairs such that one is a power multiple of other

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)

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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);   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



Output :

6

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