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 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 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 = 2³ * 3² * 5¹

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 - val₁), ...(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; 
} 

Output:

Time complexity: O(N * log²N)

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My Personal Notes

C++

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