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Find largest prime factor of a number

Given a positive integer ‘n'( 1 <= n <= 10¹⁵). Find the largest prime factor of a number.

Input: 6
Output: 3
Explanation
Prime factor of 6 are- 2, 3
Largest of them is '3'

Input: 15
Output: 5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The approach is simple, just factorise the given number by dividing it with the divisor of a number and keep updating the maximum prime factor. See this to understand more.

C++

// C++ Program to find largest prime 
// factor of number 
#include <iostream> 
#include<bits/stdc++.h> 
using namespace std; 
  
// A function to find largest prime factor 
long long maxPrimeFactors(long long n) 
{ 
    // Initialize the maximum prime factor 
    // variable with the lowest one 
    long long maxPrime = -1; 
  
    // Print the number of 2s that divide n 
    while (n % 2 == 0) { 
        maxPrime = 2; 
        n >>= 1; // equivalent to n /= 2 
    } 
  
    // n must be odd at this point, thus skip 
    // the even numbers and iterate only for 
    // odd integers 
    for (int i = 3; i <= sqrt(n); i += 2) { 
        while (n % i == 0) { 
            maxPrime = i; 
            n = n / i; 
        } 
    } 
  
    // This condition is to handle the case 
    // when n is a prime number greater than 2 
    if (n > 2) 
        maxPrime = n; 
  
    return maxPrime; 
} 
  
// Driver program to test above function 
int main() 
{ 
    long long n = 15; 
    cout << maxPrimeFactors(n) << endl; 
  
    n = 25698751364526; 
    cout <<  maxPrimeFactors(n); 
  
} 
// This code is contributed by Shivi_Aggarwal 

C

// C Program to find largest prime 
// factor of number 
#include <math.h> 
#include <stdio.h> 
  
// A function to find largest prime factor 
long long maxPrimeFactors(long long n) 
{ 
    // Initialize the maximum prime factor 
    // variable with the lowest one 
    long long maxPrime = -1; 
  
    // Print the number of 2s that divide n 
    while (n % 2 == 0) { 
        maxPrime = 2; 
        n >>= 1; // equivalent to n /= 2 
    } 
  
    // n must be odd at this point, thus skip 
    // the even numbers and iterate only for 
    // odd integers 
    for (int i = 3; i <= sqrt(n); i += 2) { 
        while (n % i == 0) { 
            maxPrime = i; 
            n = n / i; 
        } 
    } 
  
    // This condition is to handle the case 
    // when n is a prime number greater than 2 
    if (n > 2) 
        maxPrime = n; 
  
    return maxPrime; 
} 
  
// Driver program to test above function 
int main() 
{ 
    long long n = 15; 
    printf("%lld\n", maxPrimeFactors(n)); 
  
    n = 25698751364526; 
    printf("%lld", maxPrimeFactors(n)); 
  
    return 0; 
} 

Java

// Java Program to find largest 
// prime factor of number 
import java.io.*; 
import java.util.*; 
  
class GFG { 
  
    // function to find largest prime factor 
    static long maxPrimeFactors(long n) 
    { 
        // Initialize the maximum prime 
        // factor variable with the 
        // lowest one 
        long maxPrime = -1; 
  
        // Print the number of 2s 
        // that divide n 
        while (n % 2 == 0) { 
            maxPrime = 2; 
  
            // equivalent to n /= 2 
            n >>= 1; 
        } 
  
        // n must be odd at this point, 
        // thus skip the even numbers 
        // and iterate only for odd 
        // integers 
        for (int i = 3; i <= Math.sqrt(n); i += 2) { 
            while (n % i == 0) { 
                maxPrime = i; 
                n = n / i; 
            } 
        } 
  
        // This condition is to handle 
        // the case when n is a prime 
        // number greater than 2 
        if (n > 2) 
            maxPrime = n; 
  
        return maxPrime; 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        Long n = 15l; 
        System.out.println(maxPrimeFactors(n)); 
  
        n = 25698751364526l; 
        System.out.println(maxPrimeFactors(n)); 
    } 
} 
  
// This code is contributed by Gitanjali 

Python3

# Python3 code to find largest prime 
# factor of number 
import math 
  
# A function to find largest prime factor 
def maxPrimeFactors (n): 
      
    # Initialize the maximum prime factor 
    # variable with the lowest one 
    maxPrime = -1
      
    # Print the number of 2s that divide n 
    while n % 2 == 0: 
        maxPrime = 2
        n >>= 1     # equivalent to n /= 2 
          
    # n must be odd at this point,  
    # thus skip the even numbers and  
    # iterate only for odd integers 
    for i in range(3, int(math.sqrt(n)) + 1, 2): 
        while n % i == 0: 
            maxPrime = i 
            n = n / i 
      
    # This condition is to handle the  
    # case when n is a prime number  
    # greater than 2 
    if n > 2: 
        maxPrime = n 
      
    return int(maxPrime) 
  
# Driver code to test above function 
n = 15
print(maxPrimeFactors(n)) 
  
n = 25698751364526
print(maxPrimeFactors(n)) 
  
# This code is contributed by "Sharad_Bhardwaj". 

C#

// C# program to find largest 
// prime factor of number 
using System; 
  
class GFG { 
  
    // function to find largest prime factor 
    static long maxPrimeFactors(long n) 
    { 
        // Initialize the maximum prime 
        // factor variable with the 
        // lowest one 
        long maxPrime = -1; 
  
        // Print the number of 2s 
        // that divide n 
        while (n % 2 == 0) { 
            maxPrime = 2; 
  
            // equivalent to n /= 2 
            n >>= 1; 
        } 
  
        // n must be odd at this point, 
        // thus skip the even numbers 
        // and iterate only for odd 
        // integers 
        for (int i = 3; i <= Math.Sqrt(n); i += 2) { 
            while (n % i == 0) { 
                maxPrime = i; 
                n = n / i; 
            } 
        } 
  
        // This condition is to handle 
        // the case when n is a prime 
        // number greater than 2 
        if (n > 2) 
            maxPrime = n; 
  
        return maxPrime; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        long n = 15L; 
        Console.WriteLine(maxPrimeFactors(n)); 
  
        n = 25698751364526L; 
        Console.WriteLine(maxPrimeFactors(n)); 
    } 
} 
  
// This code is contributed by vt_m 

PHP

<?php 
// PHP Program to find 
// largest prime factor 
// of number 
  
// A function to find  
// largest prime factor 
function maxPrimeFactors($n) 
{ 
      
    // Initialize the maximum 
    // prime factor variable  
    // with the lowest one 
    $maxPrime = -1; 
  
    // Print the number of 
    // 2s that divide n 
    while ($n % 2 == 0) 
    { 
        $maxPrime = 2; 
          
        // equivalent to n /= 2 
        $n >>= 1;  
    } 
  
    // n must be odd at  
    // this point, thus skip 
    // the even numbers  
    // and iterate only for 
    // odd integers 
    for ($i = 3; $i <= sqrt($n); $i += 2) 
    { 
        while ($n % $i == 0) 
        { 
            $maxPrime = $i; 
            $n = $n / $i; 
        } 
    } 
  
    // This condition is  
    // to handle the case  
    // when n is a prime  
    // number greater than 2 
    if ($n > 2) 
        $maxPrime = $n; 
  
    return $maxPrime; 
} 
  
    // Driver Code 
    $n = 15; 
    echo maxPrimeFactors($n), "\n"; 
  
    $n = 25698751364526; 
    echo maxPrimeFactors($n), "\n"; 
  
// This code is contributed by aj_36 
?> 

Output:

5
328513

Time complexity: $\text{O}(\sqrt{n})$
Auxiliary space: $\text{O}(1)$

My Personal Notes

Recommended Posts:

Shubham Bansal 13

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Improved By : jit_t, Shivi_Aggarwal

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