# Count pairs in Array whose product is a Kth power of any positive integer

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)

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