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] = ZK 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 Z3
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 Z2



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

Below is the implementation of the above approach:

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// 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;
}
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Output:
5

Time complexity: O(N * log2N)

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