Distinct Prime Factors of a given number N
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:
- Create a map visited to keep track of all previous prime factors.
- Create a variable C, and initialize it with 2.
- 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++
#include <bits/stdc++.h>
using namespace std;
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++;
}
}
int main()
{
int N = 39;
distinctPrimeFactors(N);
return 0;
}
|
Java
import java.util.*;
public class GFG
{
static void distinctPrimeFactors( int N)
{
if (N < 2 ) {
System.out.print(- 1 );
}
int c = 2 ;
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++;
}
}
public static void main(String[] args)
{
int N = 39 ;
distinctPrimeFactors(N);
}
}
|
Python3
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
if __name__ = = "__main__" :
N = 39
distinctPrimeFactors(N)
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static void distinctPrimeFactors( int N)
{
if (N < 2) {
Console.Write(-1);
}
int c = 2;
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++;
}
}
public static void Main()
{
int N = 39;
distinctPrimeFactors(N);
}
}
|
Javascript
<script>
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++;
}
}
let N = 39;
distinctPrimeFactors(N);
</script>
|
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
#include <bits/stdc++.h>
using namespace std;
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;
}
int main()
{
int N = 315;
distinctPrimeFactors(N);
return 0;
}
|
Java
import java.util.*;
class GFG
{
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);
}
}
public static void main(String[] args)
{
int N = 315 ;
distinctPrimeFactors(N);
}
}
|
Python3
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)
N = 315
distinctPrimeFactors(N);
|
C#
using System;
using System.Collections.Generic;
class GFG
{
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);
}
}
public static void Main()
{
int N = 315;
distinctPrimeFactors(N);
}
}
|
Javascript
<script>
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);
}
var N = 315;
distinctPrimeFactors(N);
</script>
|
Time Complexity: O(N^(1/2))
Auxiliary Space: O(N^(1/2))
Last Updated :
04 Jan, 2023
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