Distinct Prime Factors of an Array

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
Exlpanation:
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:

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#include <bits/stdc++.h>
  
using namespace std;
  
//cppimplementation of the above approach
  
//Function to return an array
//of prime numbers upto n
//using Sieve of Eratosthenes
vector<int> seive(int n){
    vector<int> prime (n + 1,0);
    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;
      }
  
  
    vector<int> allPrimes;
    for (int i =2;i<n;i++) if (prime[i]==0) allPrimes.push_back(i);
    return allPrimes;
  }
  
//Function to return distinct
//prime factors from the given array
vector<int> distPrime(vector<int> arr, vector<int> allPrimes){
  
    //Creating an empty array
    //to store distinct prime factors
    vector<int> 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<int> allPrimes = seive(10000);
  
  vector<int> arr = {15, 30, 60};
  vector<int> ans = distPrime(arr, allPrimes);
  cout<<"[";
  for(int i:ans) cout<<i<<" ";
  cout<<"]";
  
}
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// Java implementation of the above approach
import java.util.*;
  
class GFG
{
  
// Function to return an array
// of prime numbers upto n
// using Sieve of Eratosthenes
static ArrayList<Integer> seive(int n){
    ArrayList<Integer> prime = new ArrayList<Integer>();
    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<Integer> allPrimes = new ArrayList<Integer>();
    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 array
static ArrayList<Integer> distPrime(ArrayList<Integer> arr, 
                            ArrayList<Integer> allPrimes){
  
    // Creating an empty array
    // to store distinct prime factors
    ArrayList<Integer> list1 = new ArrayList<Integer>();
  
    // 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 code
public static void main(String args[])
{
      
    // Finding prime numbers upto 10000
    // using Sieve of Eratosthenes
    ArrayList<Integer> allPrimes = new ArrayList<Integer>(seive(10000));
    ArrayList<Integer> arr = new ArrayList<Integer>();
    arr.add(15);
    arr.add(30);
    arr.add(60);
    ArrayList<Integer> ans = new ArrayList<Integer>(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
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# Python3 implementation of the above approach
  
# Function to return an array 
# of prime numbers upto n 
# using Sieve of Eratosthenes
def seive(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 array 
def 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 code
if __name__=="__main__":
  
    # Finding prime numbers upto 10000
    # using Sieve of Eratosthenes
    allPrimes = seive(10000)
  
    arr = [15, 30, 60]
    ans = distPrime(arr, allPrimes)
    print(ans)
  
# This code is contributed by mohit kumar 29
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Output:
[2, 3, 5]

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