Distinct Prime Factors of a given number N - GeeksforGeeks (2024)

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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 initialize 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
 
# python3 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=" ")
See Also
I am the Smallest Number Having Four Different Prime Factors. Can You Find me??[Solved]Prime Numbers - Definition, Chart, Prime Numbers 1 to 1000, ExamplesWhat are Prime Numbers 1 to 100? Definition, Chart, ExamplesFactors of 120 | Prime Factorization of 120, Factor Tree of 120
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;
}
if (N == 2) {
cout << 2;
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;
}
 
 
Java
 
// Java program for the above approach
import java.util.*;
class GFG 
{
// Function to find distinct prime factor
// of a number N
static void distinctPrimeFactors(int N)
{
See Also
Factors of 27? How to Find the Factors of 27 by Prime Factorization Method?
if (N < 2) {
System.out.print(-1);
return;
}
if (N == 2) {
System.out.print(2);
return;
}
HashMap<Integer, Boolean> visited = new HashMap<>();
for (int i = 2; i * i <= N; i++) {
while (N % i == 0) {
if (!visited.containsKey(i)) {
System.out.print(i + " ");
visited.put(i, true);
}
N /= i;
}
}
if (N > 2) {
System.out.print(N);
}
}
// Driver Code
public static void main(String[] args)
{
int N = 315;
distinctPrimeFactors(N);
}
}
// This code is contributed by Taranpreet
 
 
Python3
 
# Python program for the above approach
# Function to find distinct prime factor
# of a number N
def distinctPrimeFactors(N):
if (N < 2):
print(-1)
return
if N == 2:
print(2)
return
visited = {}
i = 2
while(i * i <= N):
while(N % i == 0):
if(i not in visited):
print(i , end = " ")
visited[i] = 1
N //= i
i+=1
if(N > 2):
print(N)
# Driver Code
N = 315
distinctPrimeFactors(N);
# This code is contributed by Shubham Singh
 
 
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);
return;
}
if (N == 2) {
Console.Write(2);
return;
}
Dictionary<int, bool> visited =
new Dictionary<int, bool>();
for (int i = 2; i * i <= N; i++) {
while (N % i == 0) {
if (!visited.ContainsKey(i)) {
Console.Write(i + " ");
visited[i] = true;
}
N /= i;
}
}
if (N > 2) {
Console.Write(N);
}
}
// Driver code
public static void Main()
{
int N = 315;
distinctPrimeFactors(N);
}
}
// This code is contributed by avijitmondal1998
 
 
Javascript
 
<script>
// Javascript program for the above approach
// Function to find distinct prime factor
// of a number N
function distinctPrimeFactors(N)
{
if (N < 2) {
document.write(-1);
return;
}
if (N === 2) {
document.write(2);
return;
}
visited = {};
for(var i = 2; i * i <= N; i++)
{
while(N % i == 0)
{
if(!visited[i])
{
document.write(i + " ");
visited[i] = 1;
}
N /= i;
}
}
if(N > 2)
document.write(N);
}
// Driver Code
var N = 315;
distinctPrimeFactors(N);
// This code is contributed by Shubham Singh
</script>
 
 
Output
3 5 7
Time Complexity: O(N^(1/2))
Auxiliary Space: O(N^(1/2))

 
 
 
   
 
 
 
 
 
 
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 `; }, error:function(data) { jQuery('.spinner-loading-overlay:eq(0)').remove(); jQuery('#suggestion-modal-alert').html("Something went wrong."); jQuery('#suggestion-modal-alert').show(); error_msg = true; } }); } else{ jQuery('.spinner-loading-overlay:eq(0)').remove(); jQuery('#suggestion-modal-alert').html("Minimum 5 Words and Maximum Character limit is 2000."); jQuery('#suggestion-modal-alert').show(); jQuery('#suggestion-section-textarea').focus(); error_msg = true; } } else{ jQuery('.spinner-loading-overlay:eq(0)').remove(); jQuery('#suggestion-modal-alert').html("Enter atleast four words !"); jQuery('#suggestion-modal-alert').show(); jQuery('#suggestion-section-textarea').focus(); error_msg = true; } if(error_msg){ setTimeout(() => { jQuery('#suggestion-section-textarea').focus(); jQuery('#suggestion-modal-alert').hide(); }, 3000); } } document.querySelector('.suggest-bottom-btn').addEventListener('click', function(){ jQuery('body').append('
'); jQuery('.spinner-loading-overlay').show(); if(loginData && loginData.isLoggedIn) { suggestionCall(); return; } // load the captcha script and set the token loadScript('https://www.google.com/recaptcha/api.js?render=6LdMFNUZAAAAAIuRtzg0piOT-qXCbDF-iQiUi9KY',[], function() { setGoogleRecaptcha(); }); }); $('.improvement-bottom-btn.create-improvement-btn').click(function() { //create improvement button is clicked $('body').append('
'); $('.spinner-loading-overlay').show(); // send this option via create-improvement-post api jQuery.ajax({ url: writeApiUrl + 'create-improvement-post/?v=1', type: "POST", contentType: 'application/json; charset=utf-8', dataType: 'json', xhrFields: { withCredentials: true }, data: JSON.stringify({ gfg_id: post_id }), success:function(result) { $('.spinner-loading-overlay:eq(0)').remove(); $('.improve-modal--overlay').hide(); $('.improvement-reason-modal').hide(); $('.create-improvement-redirection-to-write').attr('href',writeUrl + 'improve-post/' + `${result.id}` + '/', '_blank'); $('.create-improvement-redirection-to-write')[0].click(); }, error:function(e) { $('.spinner-loading-overlay:eq(0)').remove(); var result = e.responseJSON; if(result.detail.non_field_errors.length){ $('.improve-modal--improve-content .improve-modal--improve-content-modified').text(`${result.detail.non_field_errors}.`); jQuery('.improve-modal--overlay').show(); jQuery('.improve-modal--improvement').show(); $('.locked-status--impove-modal').css("display","block"); $('.unlocked-status--improve-modal-content').css("display","none"); $('.improve-modal--improvement').attr("status","locked"); $('.improvement-reason-modal').hide(); } }, }); }); 

                                                        Distinct Prime Factors of a given number N - GeeksforGeeks (2024)
                        

                            
                        

                    
                                                                                
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