# Count of subsets whose product is multiple of unique primes

• Difficulty Level : Expert
• Last Updated : 05 Oct, 2021

Given an array arr[] of size N,  the task is to count the number of non-empty subsets whose product is equal to P1×P2×P3×……..×Pk  where P1, P2, P3, …….Pk are distinct prime numbers.

Examples:

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Input: arr[ ] = {2, 4, 7, 10}
Output: 5
Explanation: There are a total of 5 subsets whose product is the product of distinct primes.
Subset 1: {2} -> 2
Subset 2: {2, 7} -> 2×7
Subset 3: {7} -> 7
Subset 4: {7, 10} -> 2×5×7
Subset 5: {10} -> 2×5

Input: arr[ ] = {12, 9}
Output: 0

Approach: The main idea is to find the numbers which are products of only distinct primes and call the recursion either to include them in the subset or not include in the subset. Also, an element is only added to the subset if and only if the GCD of the whole subset after adding the element is 1. Follow the steps below to solve the problem:

• Initialize a dict, say, Freq, to store the frequency of array elements.
• Initialize an array, say, Unique[] and store all those elements which are products of only distinct primes.
• Call a recursive function, say Countprime(pos, curSubset) to count all those subsets.
• Base Case: if pos equals the size of the unique array:
• if curSubset is empty, then return 0
• else, return the product of frequencies of each element of curSubset.
• Check if the element at pos can be taken in the current subset or not
• If taken, then call recursive functions as the sum of countPrime(pos+1, curSubset) and countPrime(pos+1, curSubset+[unique[pos]]).
• else, call countPrime(pos+1, curSubset).
• Print the ans returned from the function.

Below is the implementation of the above approach:

## Python3

 `# Python program for the above approach` `# Importing the module``from` `math ``import` `gcd, sqrt` `# Function to check number has``# distinct prime``def` `checkDistinctPrime(n):``    ``original ``=` `n``    ``product ``=` `1` `    ``# While N has factors of``    ``# two``    ``if` `(n ``%` `2` `=``=` `0``):``        ``product ``*``=` `2``        ``while` `(n ``%` `2` `=``=` `0``):``            ``n ``=` `n``/``/``2``    ` `    ``# Traversing till sqrt(N)``    ``for` `i ``in` `range``(``3``, ``int``(sqrt(n)), ``2``):``      ` `        ``# If N has a factor of i``        ``if` `(n ``%` `i ``=``=` `0``):``            ``product ``=` `product ``*` `i``            ` `            ``# While N has a factor``            ``# of i``            ``while` `(n ``%` `i ``=``=` `0``):``                ``n ``=` `n``/``/``i` `    ``# Covering case, N is Prime``    ``if` `(n > ``2``):``        ``product ``=` `product ``*` `n` `    ``return` `product ``=``=` `original`  `# Function to check wheather num``# can be added to the subset``def` `check(pos, subset, unique):``    ``for` `num ``in` `subset:``        ``if` `gcd(num, unique[pos]) !``=` `1``:``            ``return` `False``    ``return` `True`  `# Recursive Function to count subset``def` `countPrime(pos, currSubset, unique, frequency):` `    ``# Base Case``    ``if` `pos ``=``=` `len``(unique):``        ` `        ``# If currSubset is empty``        ``if` `not` `currSubset:``            ``return` `0` `        ``count ``=` `1``        ``for` `element ``in` `currSubset:``            ``count ``*``=` `frequency[element]``        ``return` `count` `    ``# If Unique[pos] can be added to``    ``# the Subset``    ``if` `check(pos, currSubset, unique):``        ``return` `countPrime(pos ``+` `1``, currSubset, \``                          ``unique, frequency)\``             ``+` `countPrime(pos ``+` `1``, currSubset``+``[unique[pos]], \``                          ``unique, frequency)``    ``else``:``        ``return` `countPrime(pos ``+` `1``, currSubset, \``                          ``unique, frequency)``  ` `# Function to count the subsets``def` `countSubsets(arr, N):``  ` `    ``# Initialize unique``    ``unique ``=` `set``()``    ``for` `element ``in` `arr:``        ``# Check it is a product of``        ``# distinct primes``        ``if` `checkDistinctPrime(element):``            ``unique.add(element)` `    ``unique ``=` `list``(unique)``    ` `    ``# Count frequency of unique element``    ``frequency ``=` `dict``()``    ``for` `element ``in` `unique:``        ``frequency[element] ``=` `arr.count(element)` `    ``# Function Call``    ``ans ``=` `countPrime(``0``, [], unique, frequency)``    ``return` `ans` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``# Given Input``    ``arr ``=` `[``2``, ``4``, ``7``, ``10``]``    ``N ``=` `len``(arr)``    ` `    ``# Function Call``    ``ans ``=` `countSubsets(arr, N)``    ``print``(ans)`

## Javascript

 ``
Output
`5`

Time Complexity: O(2N)
Auxiliary Space: O(N)

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