# Distinct Prime Factors of Array Product

• Difficulty Level : Medium
• Last Updated : 14 May, 2021

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 an 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.

## C++

 `// C++ program to count distinct prime` `// factors of a number.` `#include ` `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 a)` `    ``{` `        ``// use set to store distinct factors` `        ``HashSet m = ``new` `HashSet();`   `        ``// 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 a = ``new` `Vector();` `        ``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

 ``

Output :

`3`

My Personal Notes arrow_drop_up
Recommended Articles
Page :