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Python 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 (int i=0; i ≤ n-1; i++)
    {
        for (int j=i+1; j ≤ n-1; 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;

 

 

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

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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