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

• Difficulty Level : Expert
• Last Updated : 08 Apr, 2020

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 * A = 1 * 8 = 2^3
A * A = 1 * 1 = 1^3
A * A = 3 * 9 = 3^3
A * A = 9 * 24 = 6^3
A * A = 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

• 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 - valn)}
=> {(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 `` ` `#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;``    ``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 > getFact(``int` `x)``{``    ``// Prime factors along with powers``    ``vector > 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 >,``        ``int``>``        ``count_of_L;`` ` `    ``// Loop to iterate over the``    ``// elements of the array``    ``for` `(``int` `i = 0; i < n; i++) {`` ` `        ``// Factorise each element``        ``vector >``            ``factors = getFact(a[i]);``        ``sort(factors.begin(),``             ``factors.end());`` ` `        ``vector > 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 > 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:
```5
```

Time complexity: O(N * log2N)

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