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

• Last Updated : 31 May, 2022

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 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[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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