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;
// 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;
} |
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!