k-th prime factor of a given number
Given two numbers n and k, print k-th prime factor among all prime factors of n. For example, if the input number is 15 and k is 2, then output should be “5”. And if the k is 3, then output should be “-1” (there are less than k prime factors).
Examples:
Input : n = 225, k = 2
Output : 3
Prime factors are 3 3 5 5. Second
prime factor is 3.
Input : n = 81, k = 5
Output : -1
Prime factors are 3 3 3 3
A Simple Solution is to first find prime factors of n. While finding prime factors, keep track of count. If count becomes k, we return current prime factor.
C++
# include<bits/stdc++.h>
using namespace std;
int kPrimeFactor( int n, int k)
{
while (n%2 == 0)
{
k--;
n = n/2;
if (k == 0)
return 2;
}
for ( int i = 3; i <= sqrt (n); i = i+2)
{
while (n%i == 0)
{
if (k == 1)
return i;
k--;
n = n/i;
}
}
if (n > 2 && k == 1)
return n;
return -1;
}
int main()
{
int n = 12, k = 3;
cout << kPrimeFactor(n, k) << endl;
n = 14, k = 3;
cout << kPrimeFactor(n, k) << endl;
return 0;
}
|
Java
import java.io.*;
import java.math.*;
class GFG{
static int kPrimeFactor( int n, int k)
{
while (n % 2 == 0 )
{
k--;
n = n / 2 ;
if (k == 0 )
return 2 ;
}
for ( int i = 3 ; i <= Math.sqrt(n); i = i + 2 )
{
while (n % i == 0 )
{
if (k == 1 )
return i;
k--;
n = n / i;
}
}
if (n > 2 && k == 1 )
return n;
return - 1 ;
}
public static void main(String args[])
{
int n = 12 , k = 3 ;
System.out.println(kPrimeFactor(n, k));
n = 14 ; k = 3 ;
System.out.println(kPrimeFactor(n, k));
}
}
|
Python3
import math
def kPrimeFactor(n,k) :
while (n % 2 = = 0 ) :
k = k - 1
n = n / / 2
if (k = = 0 ) :
return 2
i = 3
while i < = math.sqrt(n) :
while (n % i = = 0 ) :
if (k = = 1 ) :
return i
k = k - 1
n = n / / i
i = i + 2
if (n > 2 and k = = 1 ) :
return n
return - 1
n = 12
k = 3
print (kPrimeFactor(n, k))
n = 14
k = 3
print (kPrimeFactor(n, k))
|
C#
using System;
class GFG {
static int kPrimeFactor( int n, int k)
{
while (n % 2 == 0)
{
k--;
n = n / 2;
if (k == 0)
return 2;
}
for ( int i = 3; i <= Math.Sqrt(n);
i = i + 2)
{
while (n % i == 0)
{
if (k == 1)
return i;
k--;
n = n / i;
}
}
if (n > 2 && k == 1)
return n;
return -1;
}
public static void Main()
{
int n = 12, k = 3;
Console.WriteLine(kPrimeFactor(n, k));
n = 14; k = 3;
Console.WriteLine(kPrimeFactor(n, k));
}
}
|
PHP
<?php
function kPrimeFactor( $n , $k )
{
while ( $n %2 == 0)
{
$k --;
$n = $n /2;
if ( $k == 0)
return 2;
}
for ( $i = 3; $i <= sqrt( $n ); $i = $i +2)
{
while ( $n % $i == 0)
{
if ( $k == 1)
return $i ;
$k --;
$n = $n / $i ;
}
}
if ( $n > 2 && $k == 1)
return $n ;
return -1;
}
{
$n = 12;
$k = 3;
echo kPrimeFactor( $n , $k ), "\n" ;
$n = 14;
$k = 3;
echo kPrimeFactor( $n , $k ) ;
return 0;
}
?>
|
Javascript
<script>
function kPrimeFactor(n, k)
{
while (n % 2 == 0)
{
k--;
n = n / 2;
if (k == 0)
return 2;
}
for (let i = 3; i <= Math.sqrt(n); i = i + 2)
{
while (n % i == 0)
{
if (k == 1)
return i;
k--;
n = n / i;
}
}
if (n > 2 && k == 1)
return n;
return -1;
}
let n = 12, k = 3;
document.write(kPrimeFactor(n, k) + "<br/>" );
n = 14; k = 3;
document.write(kPrimeFactor(n, k));
</script>
|
Output:
3
-1
Time Complexity: O(?n log n)
Auxiliary Space: O(1)
An Efficient Solution is to use Sieve of Eratosthenes. Note that this solution is efficient when we need k-th prime factor for multiple test cases. For a single case, previous approach is better.
The idea is to do preprocessing and store least prime factor of all numbers in given range. Once we have least prime factors stored in an array, we can find k-th prime factor by repeatedly dividing n with least prime factor while it is divisible, then repeating the process for reduced n.
C++
#include<bits/stdc++.h>
using namespace std;
const int MAX = 10001;
void sieveOfEratosthenes( int s[])
{
vector < bool > prime(MAX+1, false );
for ( int i=2; i<=MAX; i+=2)
s[i] = 2;
for ( int i=3; i<=MAX; i+=2)
{
if (prime[i] == false )
{
s[i] = i;
for ( int j=i; j*i<=MAX; j+=2)
{
if (prime[i*j] == false )
{
prime[i*j] = true ;
s[i*j] = i;
}
}
}
}
}
int kPrimeFactor( int n, int k, int s[])
{
while (n > 1)
{
if (k == 1)
return s[n];
k--;
n /= s[n];
}
return -1;
}
int main()
{
int s[MAX+1];
memset (s, -1, sizeof (s));
sieveOfEratosthenes(s);
int n = 12, k = 3;
cout << kPrimeFactor(n, k, s) << endl;
n = 14, k = 3;
cout << kPrimeFactor(n, k, s) << endl;
return 0;
}
|
Java
class GFG
{
static int MAX = 10001 ;
static void sieveOfEratosthenes( int []s)
{
boolean [] prime= new boolean [MAX+ 1 ];
for ( int i = 2 ; i <= MAX; i += 2 )
s[i] = 2 ;
for ( int i = 3 ; i <= MAX; i += 2 )
{
if (prime[i] == false )
{
s[i] = i;
for ( int j = i; j * i <= MAX; j += 2 )
{
if (prime[i * j] == false )
{
prime[i * j] = true ;
s[i * j] = i;
}
}
}
}
}
static int kPrimeFactor( int n, int k, int []s)
{
while (n > 1 )
{
if (k == 1 )
return s[n];
k--;
n /= s[n];
}
return - 1 ;
}
public static void main (String[] args)
{
int [] s = new int [MAX + 1 ];
sieveOfEratosthenes(s);
int n = 12 , k = 3 ;
System.out.println(kPrimeFactor(n, k, s));
n = 14 ;
k = 3 ;
System.out.println(kPrimeFactor(n, k, s));
}
}
|
Python3
MAX = 10001
def sieveOfEratosthenes(s):
prime = [ False for i in range ( MAX + 1 )]
for i in range ( 2 , MAX + 1 , 2 ):
s[i] = 2 ;
for i in range ( 3 , MAX , 2 ):
if (prime[i] = = False ):
s[i] = i
for j in range (i, MAX + 1 , 2 ):
if j * j> MAX :
break
if (prime[i * j] = = False ):
prime[i * j] = True
s[i * j] = i
def kPrimeFactor(n, k, s):
while (n > 1 ):
if (k = = 1 ):
return s[n]
k - = 1
n / / = s[n]
return - 1
s = [ - 1 for i in range ( MAX + 1 )]
sieveOfEratosthenes(s)
n = 12
k = 3
print (kPrimeFactor(n, k, s))
n = 14
k = 3
print (kPrimeFactor(n, k, s))
|
C#
using System;
class GFG
{
static int MAX = 10001;
static void sieveOfEratosthenes( int []s)
{
bool [] prime = new bool [MAX + 1];
for ( int i = 2; i <= MAX; i += 2)
s[i] = 2;
for ( int i = 3; i <= MAX; i += 2)
{
if (prime[i] == false )
{
s[i] = i;
for ( int j = i; j * i <= MAX; j += 2)
{
if (prime[i * j] == false )
{
prime[i * j] = true ;
s[i * j] = i;
}
}
}
}
}
static int kPrimeFactor( int n, int k, int []s)
{
while (n > 1)
{
if (k == 1)
return s[n];
k--;
n /= s[n];
}
return -1;
}
static void Main()
{
int [] s = new int [MAX + 1];
sieveOfEratosthenes(s);
int n = 12, k = 3;
Console.WriteLine(kPrimeFactor(n, k, s));
n = 14;
k = 3;
Console.WriteLine(kPrimeFactor(n, k, s));
}
}
|
PHP
<?php
$MAX = 10001;
function sieveOfEratosthenes(& $s )
{
global $MAX ;
$prime = array_fill (0, $MAX + 1, false);
for ( $i = 2; $i <= $MAX ; $i += 2)
$s [ $i ] = 2;
for ( $i = 3; $i <= $MAX ; $i += 2)
{
if ( $prime [ $i ] == false)
{
$s [ $i ] = $i ;
for ( $j = $i ; $j * $i <= $MAX ; $j += 2)
{
if ( $prime [ $i * $j ] == false)
{
$prime [ $i * $j ] = true;
$s [ $i * $j ] = $i ;
}
}
}
}
}
function kPrimeFactor( $n , $k , $s )
{
while ( $n > 1)
{
if ( $k == 1)
return $s [ $n ];
$k --;
$n = (int)( $n / $s [ $n ]);
}
return -1;
}
$s = array_fill (0, $MAX + 1, -1);
sieveOfEratosthenes( $s );
$n = 12;
$k = 3;
print (kPrimeFactor( $n , $k , $s ) . "\n" );
$n = 14;
$k = 3;
print (kPrimeFactor( $n , $k , $s ));
?>
|
Javascript
<script>
var MAX = 10001;
function sieveOfEratosthenes(s)
{
prime=Array.from({length: MAX+1}, (_, i) => false );
for (i = 2; i <= MAX; i += 2)
s[i] = 2;
for (i = 3; i <= MAX; i += 2)
{
if (prime[i] == false )
{
s[i] = i;
for (j = i; j * i <= MAX; j += 2)
{
if (prime[i * j] == false )
{
prime[i * j] = true ;
s[i * j] = i;
}
}
}
}
}
function kPrimeFactor(n , k , s)
{
while (n > 1)
{
if (k == 1)
return s[n];
k--;
n /= s[n];
}
return -1;
}
var s = Array.from({length: MAX + 1}, (_, i) => 0);
sieveOfEratosthenes(s);
var n = 12, k = 3;
document.write(kPrimeFactor(n, k, s)+ "<br>" );
n = 14;
k = 3;
document.write(kPrimeFactor(n, k, s));
</script>
|
Output:
3
-1
Time Complexity: O(n*log(log(n)))
Auxiliary Space: O(n)
Last Updated :
28 Feb, 2023
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