Given an array **arr[]** of length **N** and an integer **K**, the task is to count pairs in the array whose product is **Kth** power of a positive integer, i.e.

A[i] * A[j] = Z

^{K}for any positive integer Z.

**Examples:**

Input:arr[] = {1, 3, 9, 8, 24, 1}, K = 3

Output:5

Explanation:

There are 5 such pairs, those can be represented as Z^{3}–

A[0] * A[3] = 1 * 8 = 2^3

A[0] * A[5] = 1 * 1 = 1^3

A[1] * A[2] = 3 * 9 = 3^3

A[2] * A[4] = 9 * 24 = 6^3

A[3] * A[5] = 8 * 1 = 2^3

Input:arr[] = {7, 4, 10, 9, 2, 8, 8, 7, 3, 7}, K = 2

Output:7

Explanation:

There are 7 such pairs, those can be represented as Z^{2}

**Approach:** The key observation in this problem is for representing any number in the form of Z^{K} then powers of prime factorization of that number must be multiple of K. Below is the illustration of the steps:

- Compute the prime factorization of each number of the array and store the prime factors in the form of key-value pair in a hash-map, where the key will be a prime factor of that element and value will be the power raised to that prime factor modulus K, in the prime factorization of that number.

**For Example:**Given Element be - 360 and K = 2 Prime Factorization = 2

^{3}* 3^{2}* 5^{1}Key-value pairs for this would be, => {(2, 3 % 2), (3, 2 % 2), (5, 1 % 2)} => {(2, 1), (5, 1)} // Notice that prime number 3 // is ignored because of the // modulus value was 0 - Traverse over the array and create a frequency hash-map in which the key-value pairs would be defined as follows:
Key: Prime Factors pairs mod K Value: Frequency of this Key

- Finally, Traverse for each element of the array and check required prime factors are present in hash-map or not. If yes, then there will be
**F**number of possible pairs, where F is the frequency.**Example:**Given Number be - 360, K = 3 Prime Factorization - => {(3, 2), (5, 1)} Required Prime Factors - => {(p1, K - val

_{1}), ...(pn, K - val_{n})} => {(3, 3 - 2), (5, 3 - 1)} => {(3, 1), (5, 2)}

Below is the implementation of the above approach:

## C++

`// C++ implementation to count the ` `// pairs whose product is Kth ` `// power of some integer Z ` ` ` `#include <bits/stdc++.h> ` ` ` `#define MAXN 100005 ` ` ` `using` `namespace` `std; ` ` ` `// Smallest prime factor ` `int` `spf[MAXN]; ` ` ` `// Sieve of eratosthenes ` `// for computing primes ` `void` `sieve() ` `{ ` ` ` `int` `i, j; ` ` ` `spf[1] = 1; ` ` ` `for` `(i = 2; i < MAXN; i++) ` ` ` `spf[i] = i; ` ` ` ` ` `// Loop for markig the factors ` ` ` `// of prime number as non-prime ` ` ` `for` `(i = 2; i < MAXN; i++) { ` ` ` `if` `(spf[i] == i) { ` ` ` `for` `(j = i * 2; ` ` ` `j < MAXN; j += i) { ` ` ` `if` `(spf[j] == j) ` ` ` `spf[j] = i; ` ` ` `} ` ` ` `} ` ` ` `} ` `} ` ` ` `// Function to factorize the ` `// number N into its prime factors ` `vector<pair<` `int` `, ` `int` `> > getFact(` `int` `x) ` `{ ` ` ` `// Prime factors along with powers ` ` ` `vector<pair<` `int` `, ` `int` `> > factors; ` ` ` ` ` `// Loop while the X is not ` ` ` `// equal to 1 ` ` ` `while` `(x != 1) { ` ` ` ` ` `// Smallest prime ` ` ` `// factor of x ` ` ` `int` `z = spf[x]; ` ` ` `int` `cnt = 0; ` ` ` `// Count power of this ` ` ` `// prime factor in x ` ` ` `while` `(x % z == 0) ` ` ` `cnt++, x /= z; ` ` ` ` ` `factors.push_back( ` ` ` `make_pair(z, cnt)); ` ` ` `} ` ` ` `return` `factors; ` `} ` ` ` `// Function to count the pairs ` `int` `pairsWithKth(` `int` `a[], ` `int` `n, ` `int` `k) ` `{ ` ` ` ` ` `// Precomputation ` ` ` `// for factorisation ` ` ` `sieve(); ` ` ` ` ` `int` `answer = 0; ` ` ` ` ` `// Data structure for storing ` ` ` `// list L for each element along ` ` ` `// with frequency of occurence ` ` ` `map<vector<pair<` `int` `, ` ` ` `int` `> >, ` ` ` `int` `> ` ` ` `count_of_L; ` ` ` ` ` `// Loop to iterate over the ` ` ` `// elements of the array ` ` ` `for` `(` `int` `i = 0; i < n; i++) { ` ` ` ` ` `// Factorise each element ` ` ` `vector<pair<` `int` `, ` `int` `> > ` ` ` `factors = getFact(a[i]); ` ` ` `sort(factors.begin(), ` ` ` `factors.end()); ` ` ` ` ` `vector<pair<` `int` `, ` `int` `> > L; ` ` ` ` ` `// Loop to iterate over the ` ` ` `// factors of the element ` ` ` `for` `(` `auto` `it : factors) { ` ` ` `if` `(it.second % k == 0) ` ` ` `continue` `; ` ` ` `L.push_back( ` ` ` `make_pair( ` ` ` `it.first, ` ` ` `it.second % k)); ` ` ` `} ` ` ` ` ` `vector<pair<` `int` `, ` `int` `> > Lx; ` ` ` ` ` `// Loop to find the required prime ` ` ` `// factors for each element of array ` ` ` `for` `(` `auto` `it : L) { ` ` ` ` ` `// Represents how much remainder ` ` ` `// power needs to be added to ` ` ` `// this primes power so as to make ` ` ` `// it a multiple of k ` ` ` `Lx.push_back( ` ` ` `make_pair( ` ` ` `it.first, ` ` ` `(k - it.second + k) % k)); ` ` ` `} ` ` ` ` ` `// Add occurences of ` ` ` `// Lx till now to answer ` ` ` `answer += count_of_L[Lx]; ` ` ` ` ` `// Increment the counter for L ` ` ` `count_of_L[L]++; ` ` ` `} ` ` ` ` ` `return` `answer; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `n = 6; ` ` ` `int` `a[n] = { 1, 3, 9, 8, 24, 1 }; ` ` ` `int` `k = 3; ` ` ` ` ` `cout << pairsWithKth(a, n, k); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Output:**

5

**Time complexity:** O(N * log^{2}N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Count of index pairs in array whose range product is a positive integer
- Count of pairs in an array such that the highest power of 2 that divides their product is 1
- Count unordered pairs (i,j) such that product of a[i] and a[j] is power of two
- Count number of pairs with positive sum in an array
- Count of even and odd power pairs in an Array
- Count pairs in array such that one element is power of another
- Count pairs in a sorted array whose product is less than k
- Only integer with positive value in positive negative value in array
- Count of pairs in an array whose product is a perfect square
- Count ordered pairs of positive numbers such that their sum is S and XOR is K
- Count of pairs whose bitwise AND is a power of 2
- Minimum positive integer required to split the array equally
- Find the smallest positive integer value that cannot be represented as sum of any subset of a given array
- Pairs of Positive Negative values in an array
- Minimum product pair an array of positive Integers
- Elements of Array which can be expressed as power of some integer to given exponent K
- Count Distinct Non-Negative Integer Pairs (x, y) that Satisfy the Inequality x*x + y*y < n
- Count ordered pairs with product less than N
- Count pairs of numbers from 1 to N with Product divisible by their Sum
- Count number of ordered pairs with Even and Odd Product

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.