k-th prime factor of a given number
Last Updated :
23 Jul, 2025
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++
// Program to print kth prime factor
# include<bits/stdc++.h>
using namespace std;
// A function to generate prime factors of a
// given number n and return k-th prime factor
int kPrimeFactor(int n, int k)
{
// Find the number of 2's that divide k
while (n%2 == 0)
{
k--;
n = n/2;
if (k == 0)
return 2;
}
// n must be odd at this point. So we can skip
// one element (Note i = i +2)
for (int i = 3; i <= sqrt(n); i = i+2)
{
// While i divides n, store i and divide n
while (n%i == 0)
{
if (k == 1)
return i;
k--;
n = n/i;
}
}
// This condition is to handle the case where
// n is a prime number greater than 2
if (n > 2 && k == 1)
return n;
return -1;
}
// Driver Program
int main()
{
int n = 12, k = 3;
cout << kPrimeFactor(n, k) << endl;
n = 14, k = 3;
cout << kPrimeFactor(n, k) << endl;
return 0;
}
// JAVA Program to print kth prime factor
import java.io.*;
import java.math.*;
class GFG{
// A function to generate prime factors
// of a given number n and return k-th
// prime factor
static int kPrimeFactor(int n, int k)
{
// Find the number of 2's that
// divide k
while (n % 2 == 0)
{
k--;
n = n / 2;
if (k == 0)
return 2;
}
// n must be odd at this point.
// So we can skip one element
// (Note i = i +2)
for (int i = 3; i <= Math.sqrt(n); i = i + 2)
{
// While i divides n, store i
// and divide n
while (n % i == 0)
{
if (k == 1)
return i;
k--;
n = n / i;
}
}
// This condition is to handle the
// case where n is a prime number
// greater than 2
if (n > 2 && k == 1)
return n;
return -1;
}
// Driver Program
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));
}
}
/*This code is contributed by Nikita Tiwari.*/
# Python Program to print kth prime factor
import math
# A function to generate prime factors of a
# given number n and return k-th prime factor
def kPrimeFactor(n,k) :
# Find the number of 2's that divide k
while (n % 2 == 0) :
k = k - 1
n = n // 2
if (k == 0) :
return 2
# n must be odd at this point. So we can
# skip one element (Note i = i +2)
i = 3
while i <= math.sqrt(n) :
# While i divides n, store i and divide n
while (n % i == 0) :
if (k == 1) :
return i
k = k - 1
n = n // i
i = i + 2
# This condition is to handle the case
# where n is a prime number greater than 2
if (n > 2 and k == 1) :
return n
return -1
# Driver Program
n = 12
k = 3
print(kPrimeFactor(n, k))
n = 14
k = 3
print(kPrimeFactor(n, k))
# This code is contributed by Nikita Tiwari.
// C# Program to print kth prime factor.
using System;
class GFG {
// A function to generate prime factors
// of a given number n and return k-th
// prime factor
static int kPrimeFactor(int n, int k)
{
// Find the number of 2's that
// divide k
while (n % 2 == 0)
{
k--;
n = n / 2;
if (k == 0)
return 2;
}
// n must be odd at this point.
// So we can skip one element
// (Note i = i +2)
for (int i = 3; i <= Math.Sqrt(n);
i = i + 2)
{
// While i divides n, store i
// and divide n
while (n % i == 0)
{
if (k == 1)
return i;
k--;
n = n / i;
}
}
// This condition is to handle the
// case where n is a prime number
// greater than 2
if (n > 2 && k == 1)
return n;
return -1;
}
// Driver Program
public static void Main()
{
int n = 12, k = 3;
Console.WriteLine(kPrimeFactor(n, k));
n = 14; k = 3;
Console.WriteLine(kPrimeFactor(n, k));
}
}
// This code is contributed by nitin mittal.
<?php
// PHP Program to print kth prime factor
// A function to generate prime
// factors of a given number n
// and return k-th prime factor
function kPrimeFactor($n, $k)
{
// Find the number of 2's
// that divide k
while ($n%2 == 0)
{
$k--;
$n = $n/2;
if ($k == 0)
return 2;
}
// n must be odd at this point.
// So we can skip one element
// (Note i = i +2)
for($i = 3; $i <= sqrt($n); $i = $i+2)
{
// While i divides n,
// store i and divide n
while ($n % $i == 0)
{
if ($k == 1)
return $i;
$k--;
$n = $n / $i;
}
}
// This condition is to
// handle the case where
// n is a prime number
// greater than 2
if ($n > 2 && $k == 1)
return $n;
return -1;
}
// Driver Code
{
$n = 12;
$k = 3;
echo kPrimeFactor($n, $k),"\n" ;
$n = 14;
$k = 3;
echo kPrimeFactor($n, $k) ;
return 0;
}
// This code contributed by nitin mittal.
?>
<script>
// Javascript Program to print kth prime factor
// A function to generate prime factors
// of a given number n and return k-th
// prime factor
function kPrimeFactor(n, k)
{
// Find the number of 2's that
// divide k
while (n % 2 == 0)
{
k--;
n = n / 2;
if (k == 0)
return 2;
}
// n must be odd at this point.
// So we can skip one element
// (Note i = i +2)
for (let i = 3; i <= Math.sqrt(n); i = i + 2)
{
// While i divides n, store i
// and divide n
while (n % i == 0)
{
if (k == 1)
return i;
k--;
n = n / i;
}
}
// This condition is to handle the
// case where n is a prime number
// greater than 2
if (n > 2 && k == 1)
return n;
return -1;
}
// Driver code
let n = 12, k = 3;
document.write(kPrimeFactor(n, k) + "<br/>");
n = 14; k = 3;
document.write(kPrimeFactor(n, k));
// This code is contributed by susmitakundugoaldanga.
</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++
// C++ program to find k-th prime factor using Sieve Of
// Eratosthenes. This program is efficient when we have
// a range of numbers.
#include<bits/stdc++.h>
using namespace std;
const int MAX = 10001;
// Using SieveOfEratosthenes to find smallest prime
// factor of all the numbers.
// For example, if MAX is 10,
// s[2] = s[4] = s[6] = s[10] = 2
// s[3] = s[9] = 3
// s[5] = 5
// s[7] = 7
void sieveOfEratosthenes(int s[])
{
// Create a boolean array "prime[0..MAX]" and
// initialize all entries in it as false.
vector <bool> prime(MAX+1, false);
// Initializing smallest factor equal to 2
// for all the even numbers
for (int i=2; i<=MAX; i+=2)
s[i] = 2;
// For odd numbers less than equal to n
for (int i=3; i<=MAX; i+=2)
{
if (prime[i] == false)
{
// s(i) for a prime is the number itself
s[i] = i;
// For all multiples of current prime number
for (int j=i; j*i<=MAX; j+=2)
{
if (prime[i*j] == false)
{
prime[i*j] = true;
// i is the smallest prime factor for
// number "i*j".
s[i*j] = i;
}
}
}
}
}
// Function to generate prime factors and return its
// k-th prime factor. s[i] stores least prime factor
// of i.
int kPrimeFactor(int n, int k, int s[])
{
// Keep dividing n by least prime factor while
// either n is not 1 or count of prime factors
// is not k.
while (n > 1)
{
if (k == 1)
return s[n];
// To keep track of count of prime factors
k--;
// Divide n to find next prime factor
n /= s[n];
}
return -1;
}
// Driver Program
int main()
{
// s[i] is going to store prime factor
// of i.
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 program to find k-th prime factor
// using Sieve Of Eratosthenes. This
// program is efficient when we have
// a range of numbers.
class GFG
{
static int MAX = 10001;
// Using SieveOfEratosthenes to find smallest prime
// factor of all the numbers.
// For example, if MAX is 10,
// s[2] = s[4] = s[6] = s[10] = 2
// s[3] = s[9] = 3
// s[5] = 5
// s[7] = 7
static void sieveOfEratosthenes(int []s)
{
// Create a boolean array "prime[0..MAX]" and
// initialize all entries in it as false.
boolean[] prime=new boolean[MAX+1];
// Initializing smallest factor equal to 2
// for all the even numbers
for (int i = 2; i <= MAX; i += 2)
s[i] = 2;
// For odd numbers less than equal to n
for (int i = 3; i <= MAX; i += 2)
{
if (prime[i] == false)
{
// s(i) for a prime is the number itself
s[i] = i;
// For all multiples of current prime number
for (int j = i; j * i <= MAX; j += 2)
{
if (prime[i * j] == false)
{
prime[i * j] = true;
// i is the smallest prime factor for
// number "i*j".
s[i * j] = i;
}
}
}
}
}
// Function to generate prime factors
// and return its k-th prime factor.
// s[i] stores least prime factor of i.
static int kPrimeFactor(int n, int k, int []s)
{
// Keep dividing n by least
// prime factor while either
// n is not 1 or count of
// prime factors is not k.
while (n > 1)
{
if (k == 1)
return s[n];
// To keep track of count of prime factors
k--;
// Divide n to find next prime factor
n /= s[n];
}
return -1;
}
// Driver code
public static void main (String[] args)
{
// s[i] is going to store prime factor
// of i.
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));
}
}
// This code is contributed by mits
# python3 program to find k-th prime factor using Sieve Of
# Eratosthenes. This program is efficient when we have
# a range of numbers.
MAX = 10001
# Using SieveOfEratosthenes to find smallest prime
# factor of all the numbers.
# For example, if MAX is 10,
# s[2] = s[4] = s[6] = s[10] = 2
# s[3] = s[9] = 3
# s[5] = 5
# s[7] = 7
def sieveOfEratosthenes(s):
# Create a boolean array "prime[0..MAX]" and
# initialize all entries in it as false.
prime=[False for i in range(MAX+1)]
# Initializing smallest factor equal to 2
# for all the even numbers
for i in range(2,MAX+1,2):
s[i] = 2;
# For odd numbers less than equal to n
for i in range(3,MAX,2):
if (prime[i] == False):
# s(i) for a prime is the number itself
s[i] = i
# For all multiples of current prime number
for j in range(i,MAX+1,2):
if j*j> MAX:
break
if (prime[i*j] == False):
prime[i*j] = True
# i is the smallest prime factor for
# number "i*j".
s[i*j] = i
# Function to generate prime factors and return its
# k-th prime factor. s[i] stores least prime factor
# of i.
def kPrimeFactor(n, k, s):
# Keep dividing n by least prime factor while
# either n is not 1 or count of prime factors
# is not k.
while (n > 1):
if (k == 1):
return s[n]
# To keep track of count of prime factors
k-=1
# Divide n to find next prime factor
n //= s[n]
return -1
# Driver Program
# s[i] is going to store prime factor
# of i.
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))
# This code is contributed by mohit kumar 29
// C# program to find k-th prime factor
// using Sieve Of Eratosthenes. This
// program is efficient when we have
// a range of numbers and we
using System;
class GFG
{
static int MAX = 10001;
// Using SieveOfEratosthenes to find
// smallest prime factor of all the
// numbers. For example, if MAX is 10,
// s[2] = s[4] = s[6] = s[10] = 2
// s[3] = s[9] = 3
// s[5] = 5
// s[7] = 7
static void sieveOfEratosthenes(int []s)
{
// Create a boolean array "prime[0..MAX]"
// and initialize all entries in it as false.
bool[] prime = new bool[MAX + 1];
// Initializing smallest factor equal
// to 2 for all the even numbers
for (int i = 2; i <= MAX; i += 2)
s[i] = 2;
// For odd numbers less than equal to n
for (int i = 3; i <= MAX; i += 2)
{
if (prime[i] == false)
{
// s(i) for a prime is the
// number itself
s[i] = i;
// For all multiples of current
// prime number
for (int j = i; j * i <= MAX; j += 2)
{
if (prime[i * j] == false)
{
prime[i * j] = true;
// i is the smallest prime factor
// for number "i*j".
s[i * j] = i;
}
}
}
}
}
// Function to generate prime factors
// and return its k-th prime factor.
// s[i] stores least prime factor of i.
static int kPrimeFactor(int n, int k, int []s)
{
// Keep dividing n by least prime
// factor while either n is not 1
// or count of prime factors is not k.
while (n > 1)
{
if (k == 1)
return s[n];
// To keep track of count of
// prime factors
k--;
// Divide n to find next prime factor
n /= s[n];
}
return -1;
}
// Driver Code
static void Main()
{
// s[i] is going to store prime
// factor of i.
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));
}
}
// This code is contributed by mits
<?php
// PHP program to find k-th prime factor
// using Sieve Of Eratosthenes. This program
// is efficient when we have a range of numbers.
$MAX = 10001;
// Using SieveOfEratosthenes to find
// smallest prime factor of all the numbers.
// For example, if MAX is 10,
// s[2] = s[4] = s[6] = s[10] = 2
// s[3] = s[9] = 3
// s[5] = 5
// s[7] = 7
function sieveOfEratosthenes(&$s)
{
global $MAX;
// Create a boolean array "prime[0..MAX]"
// and initialize all entries in it as false.
$prime = array_fill(0, $MAX + 1, false);
// Initializing smallest factor equal
// to 2 for all the even numbers
for ($i = 2; $i <= $MAX; $i += 2)
$s[$i] = 2;
// For odd numbers less than equal to n
for ($i = 3; $i <= $MAX; $i += 2)
{
if ($prime[$i] == false)
{
// s(i) for a prime is the
// number itself
$s[$i] = $i;
// For all multiples of current
// prime number
for ($j = $i; $j * $i <= $MAX; $j += 2)
{
if ($prime[$i * $j] == false)
{
$prime[$i * $j] = true;
// i is the smallest prime
// factor for number "i*j".
$s[$i * $j] = $i;
}
}
}
}
}
// Function to generate prime factors and
// return its k-th prime factor. s[i] stores
// least prime factor of i.
function kPrimeFactor($n, $k, $s)
{
// Keep dividing n by least prime
// factor while either n is not 1
// or count of prime factors is not k.
while ($n > 1)
{
if ($k == 1)
return $s[$n];
// To keep track of count of
// prime factors
$k--;
// Divide n to find next prime factor
$n = (int)($n / $s[$n]);
}
return -1;
}
// Driver Code
// s[i] is going to store prime
// factor of i.
$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));
// This code is contributed by chandan_jnu
?>
<script>
// Javascript program to find k-th prime factor
// using Sieve Of Eratosthenes. This
// program is efficient when we have
// a range of numbers.
var MAX = 10001;
// Using SieveOfEratosthenes to find smallest prime
// factor of all the numbers.
// For example, if MAX is 10,
// s[2] = s[4] = s[6] = s[10] = 2
// s[3] = s[9] = 3
// s[5] = 5
// s[7] = 7
function sieveOfEratosthenes(s)
{
// Create a boolean array "prime[0..MAX]" and
// initialize all entries in it as false.
prime=Array.from({length: MAX+1}, (_, i) => false);
// Initializing smallest factor equal to 2
// for all the even numbers
for (i = 2; i <= MAX; i += 2)
s[i] = 2;
// For odd numbers less than equal to n
for (i = 3; i <= MAX; i += 2)
{
if (prime[i] == false)
{
// s(i) for a prime is the number itself
s[i] = i;
// For all multiples of current prime number
for (j = i; j * i <= MAX; j += 2)
{
if (prime[i * j] == false)
{
prime[i * j] = true;
// i is the smallest prime factor for
// number "i*j".
s[i * j] = i;
}
}
}
}
}
// Function to generate prime factors
// and return its k-th prime factor.
// s[i] stores least prime factor of i.
function kPrimeFactor(n , k , s)
{
// Keep dividing n by least
// prime factor while either
// n is not 1 or count of
// prime factors is not k.
while (n > 1)
{
if (k == 1)
return s[n];
// To keep track of count of prime factors
k--;
// Divide n to find next prime factor
n /= s[n];
}
return -1;
}
// Driver code
// s[i] is going to store prime factor
// of i.
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));
// This code contributed by shikhasingrajput
</script>
Output:
3
-1
Time Complexity: O(n*log(log(n)))
Auxiliary Space: O(n)
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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