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 = 3Output:5Explanation:

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 = 2Output:7Explanation:

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++ 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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