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# Distinct Prime Factors of a given number N

• Last Updated : 04 Jan, 2023

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=" ")
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)
{
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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