k-th prime factor of a given number - GeeksforGeeks

24 Jan, 2022
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++
Java
Python3
C#
PHP
Javascript
// 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 factorint 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 Programint 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 factorimport 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 factorimport math # A function to generate prime factors of a# given number n and return k-th prime factordef 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 Programn = 12k = 3print(kPrimeFactor(n, k)) n = 14k = 3print(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 factorfunction 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
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++
Java
Python3
C#
PHP
Javascript
// 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] = 7void 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 then 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 Programint 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] = 7static 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 then 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 codepublic 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] = 7def 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 then 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 = 12k = 3print(kPrimeFactor(n, k, s)) n = 14k = 3print(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 weusing 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] = 7static 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 then 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 Codestatic 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. $MAX = 10001; // Using SieveOfEratosthenes to find// smallest prime factor of all the numbers.// For example, if MAX is 10,// s[2] = s[4] = s[6] = s[10] = 2// s[3] = s[9] = 3// s[5] = 5// s[7] = 7function sieveOfEratosthenes(&$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 then 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] = 7function 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 then 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
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nitin mittal
Mithun Kumar
Chandan_Kumar
mohit kumar 29
susmitakundugoaldanga
shikhasingrajput
gabaa406
amartyaghoshgfg
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