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 |
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!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!