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Distinct Prime Factors of a given number N

  • Last Updated : 21 Jan, 2022

Given a number N, the task is to find the distinct Prime Factors of N.

Examples:

Input: N = 12
Output: 2 3
Explanation: The factors of 12 are 1, 2, 3, 4, 6, 12.
Among these the distinct prime factors are 2 and 3.

Input: N = 39
Output: 3 13

 

Approach: The approach is to use a map to check whether a given factor of the number has occurred earlier or not. Now follow the below steps to solve this problem:

  1. Create a map visited to keep track of all previous prime factors.
  2. Create a variable C, and initialise it with 2.
  3. While N is divisible by C, print C if C is not present in the map. Now divide N by C. Also increment C by 1.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find distinct prime factor
// of a number N
void distinctPrimeFactors(int N)
{
    if (N < 2) {
        cout << -1;
    }
 
    int c = 2;
    unordered_map<int, bool> visited;
 
    while (N > 1) {
        if (N % c == 0) {
            if (!visited) {
                cout << c << " ";
                visited = 1;
            }
            N /= c;
        }
        else
            c++;
    }
}
 
// Driver Code
int main()
{
    int N = 39;
    distinctPrimeFactors(N);
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to find distinct prime factor
  // of a number N
  static void distinctPrimeFactors(int N)
  {
    if (N < 2) {
      System.out.print(-1);
    }
 
    int c = 2;
 
    // Create a new dictionary of
    // strings, with string keys.
    HashMap<Integer, Boolean> visited = new HashMap<>();
 
    for(int i = 0; i < N; i++) {
      visited.put(i, false);
    }
 
    while (N > 1) {
      if (N % c == 0) {
        if(visited.containsKey(c)){
          if (!visited.get(c)) {
            System.out.print(c + " ");
            visited.put(c, true);
          }
        }
        N /= c;
      }
      else
        c++;
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 39;
    distinctPrimeFactors(N);
  }
}
 
// This code is contributed by Samim Hossain Mondal

Python3




# pythn3 program for the above approach
 
# Function to find distinct prime factor
# of a number N
def distinctPrimeFactors(N):
 
    if (N < 2):
        print(-1)
 
    c = 2
    visited = {}
 
    while (N > 1):
        if (N % c == 0):
            if (not c in visited):
                print(c, end=" ")
                visited = 1 if c in visited else 0
 
            N //= c
 
        else:
            c += 1
 
# Driver Code
if __name__ == "__main__":
 
    N = 39
    distinctPrimeFactors(N)
 
# This code is contributed by rakeshsahni

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
 
  // Function to find distinct prime factor
  // of a number N
  static void distinctPrimeFactors(int N)
  {
    if (N < 2) {
      Console.Write(-1);
    }
 
    int c = 2;
 
    // Create a new dictionary of
    // strings, with string keys.
    Dictionary<int, bool> visited =
      new Dictionary<int, bool>();
 
    for(int i = 0; i < N; i++) {
      visited[i] = false;
    }
 
    while (N > 1) {
      if (N % c == 0) {
        if(visited.ContainsKey(c)){
          if (!visited) {
            Console.Write(c + " ");
            visited = true;
          }
        }
        N /= c;
      }
      else
        c++;
    }
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 39;
    distinctPrimeFactors(N);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
    // JavaScript program for the above approach
 
    // Function to find distinct prime factor
    // of a number N
    const distinctPrimeFactors = (N) => {
        if (N < 2) {
            document.write(-1);
        }
 
        let c = 2;
        let visited = {};
 
        while (N > 1) {
            if (N % c == 0) {
                if (!(c in visited)) {
                    document.write(`${c} `);
                    visited = 1;
                }
                N = parseInt(N / c);
            }
            else
                c++;
        }
    }
 
    // Driver Code
 
    let N = 39;
    distinctPrimeFactors(N);
 
// This code is contributed by rakeshsahni
 
</script>
Output
3 13 

Time Complexity: O(N)
Auxiliary Space: O(N1/2)

Efficient Approach: This approach is similar to above approach where we find prime factors. The only difference is that we traverse from 2 to sqrt(n) to find all prime factors since we know that is sufficient for checking for prime numbers as well. If the number is still found to be greater than 2 then it is prime and we need to print it as well.

C++14




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find distinct prime factor
// of a number N
void distinctPrimeFactors(int N)
{
    if (N < 2) {
        cout << -1;
        return;
    }
     
    unordered_map<int, bool> visited;
     
    for(int i=2;i*i<=N;i++)
    {
        while(N%i==0)
        {
            if(!visited[i])
            {
                cout<<i<<" ";
                visited[i]=1;
            }
            N/=i;
        }
    }
    if(N>2)
    cout<<N;
}
 
// Driver Code
int main()
{
    int N = 315;
    distinctPrimeFactors(N);
    return 0;
}

Time Complexity: O(N^(1/2))

Auxiliary Space: O(N^(1/2))


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