Distinct Prime Factors of Array Product

Given an array of integers. Let us say P is the product of elements of the array. Find the number of distinct prime factors of product P.

Examples:

Input : 1 2 3 4 5 
Output : 3 
Explanation: Here P = 1 * 2 * 3 * 4 * 5 = 120. Distinct prime divisors of 120 are 2, 3 and 5. So, the output is 3.

Input : 21 30 15 24 16 
Output : 4 
Explanation: Here P = 21 * 30 * 15 * 24 * 16 = 3628800. Distinct prime divisors of 3628800 are 2, 3, 5 and 7. So, the output is 4.

Naive Approach: The simple solution for the problem would be to multiply every number in the array and then find the number of distinct prime factors of the product. 
But this method can lead to integer overflow.

Better Approach: To avoid the overflow instead of multiplying the numbers we can find the prime factors of each element separately and store the prime factors in a set or a map for unique factors.

Implementation:

C++

// C++ program to count distinct prime
// factors of a number.
#include <bits/stdc++.h>
using namespace std;

// Function to count the number of distinct prime
// factors of product of array
int Distinct_Prime_factors(vector<int> a)
{
    // use set to store distinct factors
    unordered_set<int> m;

    // iterate over every element of array
    for (int i = 0; i < a.size(); i++) {
        int sq = sqrt(a[i]);

        // from 2 to square root of number run
        // a loop and check the numbers which
        // are factors.
        for (int j = 2; j <= sq; j++) {
            if (a[i] % j == 0) {

                // if j is a factor store it in the set
                m.insert(j);

                // divide the number with j till it
                // is divisible so that only prime factors
                // are stored
                while (a[i] % j == 0) {
                    a[i] /= j;
                }
            }
        }

        // if the number is still greater than 1 then
        // it is a prime factor, insert in set
        if (a[i] > 1) {
            m.insert(a[i]);
        }
    }

    // the number of unique prime factors will
    // the size of the set
    return m.size();
}

// Driver Function
int main()
{
    vector<int> a = { 1, 2, 3, 4, 5 };
    cout << Distinct_Prime_factors(a) << '\n';
    return 0;
}

Java

// Java program to count distinct
// prime factors of a number.
import java.util.*;

class GFG {

    // Function to count the number
    // of distinct prime factors of
    // product of array
    static int Distinct_Prime_factors(Vector<Integer> a)
    {
        // use set to store distinct factors
        HashSet<Integer> m = new HashSet<Integer>();

        // iterate over every element of array
        for (int i = 0; i < a.size(); i++) {
            int sq = (int)Math.sqrt(a.get(i));

            // from 2 to square root of number
            // run a loop and check the numbers
            // which are factors.
            for (int j = 2; j <= sq; j++) {
                if (a.get(i) % j == 0) {

                    // if j is a factor store
                    // it in the set
                    m.add(j);

                    // divide the number with j
                    // till it is divisible so
                    // that only prime factors
                    // are stored
                    while (a.get(i) % j == 0) {
                        a.set(i, a.get(i) / j);
                    }
                }
            }

            // if the number is still greater
            // than 1 then it is a prime factor,
            // insert in set
            if (a.get(i) > 1) {
                m.add(a.get(i));
            }
        }

        // the number of unique prime
        // factors will the size of the set
        return m.size();
    }

    // Driver Code
    public static void main(String args[])
    {
        Vector<Integer> a = new Vector<Integer>();
        a.add(1);
        a.add(2);
        a.add(3);
        a.add(4);
        a.add(5);
        System.out.println(Distinct_Prime_factors(a));
    }
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to count distinct
# prime factors of a number
import math

# Function to count the number of distinct
# prime factors of product of array
def Distinct_Prime_factors( a):

    # use set to store distinct factors
    m = []

    # iterate over every element of array
    for i in range (len(a)) :
        sq = int(math.sqrt(a[i]))

        # from 2 to square root of number run
        # a loop and check the numbers which
        # are factors.
        for j in range(2, sq + 1) :
            if (a[i] % j == 0) :

                # if j is a factor store
                # it in the set
                m.append(j)

                # divide the number with j till it
                # is divisible so that only prime
                # factors are stored
                while (a[i] % j == 0) :
                    a[i] //= j

        # if the number is still greater
        # than 1 then it is a prime factor,
        # insert in set
        if (a[i] > 2) :
            m.append(a[i])

    # the number of unique prime factors
    # will the size of the set
    return len(m)

# Driver Code
if __name__ == "__main__":

    a = [ 1, 2, 3, 4, 5 ]
    print (Distinct_Prime_factors(a))

# This code is contributed by ita_c

C#

// C# program to count distinct
// prime factors of a number.
using System;
using System.Collections.Generic;

class GFG {

    // Function to count the number
    // of distinct prime factors of
    // product of array
    static int Distinct_Prime_factors(List<int> a)
    {
        // use set to store distinct factors
        HashSet<int> m = new HashSet<int>();

        // iterate over every element of array
        for (int i = 0; i < a.Count; i++) {
            int sq = (int)Math.Sqrt(a[i]);

            // from 2 to square root of number
            // run a loop and check the numbers
            // which are factors.
            for (int j = 2; j <= sq; j++) {
                if (a[i] % j == 0) {

                    // if j is a factor store
                    // it in the set
                    m.Add(j);

                    // divide the number with j
                    // till it is divisible so
                    // that only prime factors
                    // are stored
                    while (a[i] % j == 0) {
                        a[i] = a[i] / j;
                    }
                }
            }

            // if the number is still greater
            // than 1 then it is a prime factor,
            // insert in set
            if (a[i] > 1) {
                m.Add(a[i]);
            }
        }

        // the number of unique prime
        // factors will the size of the set
        return m.Count;
    }

    // Driver Code
    public static void Main()
    {
        List<int> a = new List<int>();
        a.Add(1);
        a.Add(2);
        a.Add(3);
        a.Add(4);
        a.Add(5);
        Console.WriteLine(Distinct_Prime_factors(a));
    }
}

// This code is contributed by ihritik

Javascript

<script>

// Javascript program to count distinct
// prime factors of a number.

// Function to count the number
// of distinct prime factors of
// product of array
function Distinct_Prime_factors(a)
{

    // Use set to store distinct factors
    let m = new Set();

    // Iterate over every element of array
    for(let i = 0; i < a.length; i++)
    {
        let sq = Math.floor(Math.sqrt(a[i]));

        // From 2 to square root of number
        // run a loop and check the numbers
        // which are factors.
        for(let j = 2; j <= sq; j++)
        {
            if (a[i] % j == 0)
            {

                // If j is a factor store
                // it in the set
                m.add(j);

                // Divide the number with j
                // till it is divisible so
                // that only prime factors
                // are stored
                while (a[i] % j == 0)
                {
                    a[i]= Math.floor(a[i] / j);
                }
            }
        }

        // If the number is still greater
        // than 1 then it is a prime factor,
        // insert in set
        if (a[i] > 1)
        {
            m.add(a[i]);
        }
    }

    // The number of unique prime
    // factors will the size of the set
    return m.size;
}

// Driver Code
let a = [ 1, 2, 3, 4, 5 ];
document.write(Distinct_Prime_factors(a));

// This code is contributed by avanitrachhadiya2155

</script>

Output

3

Time Complexity: O(n*sqrt(n))
Auxiliary Space: O(n)