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 arrayint 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 Functionint 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 numberimport math# Function to count the number of distinct # prime factors of product of arraydef 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 Codeif __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 arrayfunction 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 Codelet 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)
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