# Distinct Prime Factors of an Array

• Difficulty Level : Medium
• Last Updated : 21 Jan, 2022

Given an array arr[] of size N, the task is to find the distinct prime factors of all the numbers in the given array.

Examples:

Input: N = 3, arr[] = {12, 15, 18}
Output: 2 3 5
Explanation:
12 = 2 x 2 x 3
15 = 3 x 5
18 = 2 x 3 x 3
Distinct prime factors among the given numbers are 2, 3, 5.

Input: N = 9, arr[] = {2, 3, 4, 5, 6, 7, 8, 9, 10}
Output: 2 3 5 7

Naive Approach: A simple approach of this problem will be finding the prime factors of each number in the array. Then find the distinct prime numbers among these prime factors.
Time Complexity: O(N2)

Efficient Approach: An efficient approach is to first find all prime numbers up to the given limit using Sieve of Eratosthenes and store them in an array. For every prime number in the prime array, check if any number in the input array is divisible or not. If it is divisible, then store that prime number in the answer array. Finally, return the answer array after repeating this process for all the numbers in the given input array.

Below is the implementation of the above approach:

## C++14

 #include  using namespace std; //cppimplementation of the above approach //Function to return an array//of prime numbers upto n//using Sieve of Eratosthenesvector sieve(int n){    vector prime (n + 1,0);    int p = 2;    while(p * p<= n){        if(prime[p]== 0){            for (int i=2*p;i allPrimes;    for (int i =2;i distPrime(vector arr, vector allPrimes){     //Creating an empty array    //to store distinct prime factors    vector list1;     //Iterating through all the    //prime numbers and check if    //any of the prime numbers is a    //factor of the given input array    for (int i : allPrimes){        for (int j :arr){            if(j % i == 0){                list1.push_back(i);                break;              }            }          }    return list1;  } //Driver code int main(){  //Finding prime numbers upto 10000  //using Sieve of Eratosthenes  vector allPrimes = sieve(10000);   vector arr = {15, 30, 60};  vector ans = distPrime(arr, allPrimes);  cout<<"[";  for(int i:ans) cout<

## Java

 // Java implementation of the above approachimport java.util.*; class GFG{ // Function to return an array// of prime numbers upto n// using Sieve of Eratosthenesstatic ArrayList sieve(int n){    ArrayList prime = new ArrayList();    for(int i = 0; i < n + 1; i++)    prime.add(0);    int p = 2;    while(p * p <= n){        if(prime.get(p) == 0){            for (int i = 2 * p; i < n + 1; i += p)                prime.set(i, 1);            }        p += 1;    }     ArrayList allPrimes = new ArrayList();    for (int i = 2; i < n; i++){    if (prime.get(i) == 0)        allPrimes.add(i);    }    return allPrimes;} // Function to return distinct// prime factors from the given arraystatic ArrayList distPrime(ArrayList arr,                            ArrayList allPrimes){     // Creating an empty array    // to store distinct prime factors    ArrayList list1 = new ArrayList();     // Iterating through all the    // prime numbers and check if    // any of the prime numbers is a    // factor of the given input array    for (int i = 0; i < allPrimes.size(); i++){        for (int j = 0; j < arr.size(); j++){            if(arr.get(j) % allPrimes.get(i) == 0){                list1.add(allPrimes.get(i));                break;            }            }        }    return list1;} // Driver codepublic static void main(String args[]){         // Finding prime numbers upto 10000    // using Sieve of Eratosthenes    ArrayList allPrimes = new ArrayList(sieve(10000));    ArrayList arr = new ArrayList();    arr.add(15);    arr.add(30);    arr.add(60);    ArrayList ans = new ArrayList(distPrime(arr, allPrimes));    System.out.print("[");    for(int i = 0; i < ans.size(); i++)    System.out.print(ans.get(i) + " ");    System.out.print("]");}} // This code is contributed by Surendra_Gangwar

## Python3

 # Python3 implementation of the above approach # Function to return an array# of prime numbers upto n# using Sieve of Eratosthenesdef sieve(n):    prime =[True]*(n + 1)    p = 2    while(p * p<= n):        if(prime[p] == True):            for i in range(p * p, n + 1, p):                prime[i] = False        p += 1    allPrimes = [x for x in range(2, n)if prime[x]]    return allPrimes # Function to return distinct# prime factors from the given arraydef distPrime(arr, allPrimes):     # Creating an empty array    # to store distinct prime factors    list1 = list()         # Iterating through all the    # prime numbers and check if    # any of the prime numbers is a    # factor of the given input array    for i in allPrimes:        for j in arr:            if(j % i == 0):                list1.append(i)                break    return list1 # Driver codeif __name__=="__main__":     # Finding prime numbers upto 10000    # using Sieve of Eratosthenes    allPrimes = sieve(10000)     arr = [15, 30, 60]    ans = distPrime(arr, allPrimes)    print(ans) # This code is contributed by mohit kumar 29

## C#

 // C# implementation of the above approachusing System;using System.Collections.Generic;class GFG{ // Function to return an array// of prime numbers upto n// using Sieve of Eratosthenesstatic List sieve(int n){    List prime = new List();    for(int i = 0; i < n + 1; i++)    prime.Add(0);    int p = 2;    while(p * p <= n)    {        if(prime[p] == 0)        {            for (int i = 2 * p; i < n + 1; i += p)                prime[i]= 1;            }        p += 1;    }     List allPrimes = new List();    for (int i = 2; i < n; i++){    if (prime[i] == 0)        allPrimes.Add(i);    }    return allPrimes;} // Function to return distinct// prime factors from the given arraystatic List distPrime(List arr,                            List allPrimes){     // Creating an empty array    // to store distinct prime factors       List list1 = new List();     // Iterating through all the    // prime numbers and check if    // any of the prime numbers is a    // factor of the given input array    for (int i = 0; i < allPrimes.Count; i++){        for (int j = 0; j < arr.Count; j++){            if(arr[j] % allPrimes[i] == 0){                list1.Add(allPrimes[i]);                break;            }            }        }    return list1;} // Driver codepublic static void Main(string []args){         // Finding prime numbers upto 10000    // using Sieve of Eratosthenes    List allPrimes = new List(sieve(10000));    List arr = new List();    arr.Add(15);    arr.Add(30);    arr.Add(60);    List ans = new List(distPrime(arr, allPrimes));    Console.Write("[");    for(int i = 0; i < ans.Count; i++)    Console.Write(ans[i] + " ");    Console.Write("]");}} // This code is contributed by chitranayal

## Javascript



Output:

[2, 3, 5]

Auxiliary Space: O(10000 + |arr|)

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