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

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  • Last Updated : 31 May, 2022
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You are given an array A[] of n-elements and a positive integer k (k > 1). 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 (i = 0 to n-1)
    {
        for (j = i+1 to n-1)
        {
            // count Aj such that Ai*k^x = Aj
            int x = 0;

            // increase x till Ai * k^x lesser than 
            // 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;

 

 

C++




// 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;
}

Output : 

6

 

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

Please refer complete article on Pairs such that one is a power multiple of other for more details!


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