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

Method 1:

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

Below is the implementation of the above approach

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
     while (n % 3 == 0) {
        maxPrime = 3;
        n=n/3;
    }
 
    // now we have to iterate only for integers
    // who does not have prime factor 2 and 3
    for (int i = 5; i <= sqrt(n); i += 6) {
        while (n % i == 0) {
            maxPrime = i;
            n = n / i;
        }
      while (n % (i+2) == 0) {
            maxPrime = i+2;
            n = n / (i+2);
        }
    }
 
    // This condition is to handle the case
    // when n is a prime number greater than 4
    if (n > 4)
        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);
 
}

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
    while (n % 3 == 0) {
        maxPrime = 3;
        n = n / 3;
    }
 
    // now we have to iterate only for integers
    // who does not have prime factor 2 and 3
    for (int i = 5; i*i<=n; i += 6) {
        while (n % i == 0) {
            maxPrime = i;
            n = n / i;
        }
        while (n % (i + 2) == 0) {
            maxPrime = i + 2;
            n = n / (i + 2);
        }
    }
 
    // This condition is to handle the case
    // when n is a prime number greater than 4
    if (n > 4)
        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
        while (n % 3 == 0) {
            maxPrime = 3;
            n = n / 3;
        }
 
        // now we have to iterate only for integers
        // who does not have prime factor 2 and 3
        for (int i = 5; i <= Math.sqrt(n); i += 6) {
            while (n % i == 0) {
                maxPrime = i;
                n = n / i;
            }
            while (n % (i + 2) == 0) {
                maxPrime = i + 2;
                n = n / (i + 2);
            }
        }
 
        // This condition is to handle the case
        // when n is a prime number greater than 4
        if (n > 4)
            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));
    }
}

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
    while n % 3 == 0:
        maxPrime = 3
        n=n/3
     
    # now we have to iterate only for integers
    # who does not have prime factor 2 and 3
    for i in range(5, int(math.sqrt(n)) + 1, 6):
        while n % i == 0:
            maxPrime = i
            n = n / i
        while n % (i+2) == 0:
            maxPrime = i+2
            n = n / (i+2)
         
    # This condition is to handle the
    # case when n is a prime number
    # greater than 4
    if n > 4:
        maxPrime = n
     
    return int(maxPrime)
 
# Driver code to test above function
n = 15
print(maxPrimeFactors(n))
 
n = 25698751364526
print(maxPrimeFactors(n))

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
        while (n % 3 == 0) {
            maxPrime = 3;
            n = n / 3;
        }
        // now we have to iterate only for integers
        // who does not have prime factor 2 and 3
        for (int i = 5; i <= Math.Sqrt(n); i += 6) {
            while (n % i == 0) {
                maxPrime = i;
                n = n / i;
            }
            while (n % (i + 2) == 0) {
                maxPrime = i + 2;
                n = n / (i + 2);
            }
        }
 
        // This condition is to handle the case
        // when n is a prime number greater than 4
        if (n > 4)
            maxPrime = n;
 
        return maxPrime;
    }
 
    // Driver code
    public static void Main()
    {
        long n = 15L;
        Console.WriteLine(maxPrimeFactors(n));
 
        n = 25698751364526L;
        Console.WriteLine(maxPrimeFactors(n));
    }
}

Javascript

<script>
// Javascript program to find largest prime factor
function maxPrimeFactor(n) {
    let maxPrime = -1;
    while(n % 2 == 0) {
        n = n / 2;
        maxPrime = 2;
    }
 
    while(n % 3 == 0) {
        n = n / 3;
        maxPrime = 3;
    }
 
    for (let i = 5; i <= Math.sqrt(n); i += 6) {
        while (n % i == 0) {
            maxPrime = i;
            n = n / i;
        }
        while (n % (i + 2) == 0) {
            maxPrime = i + 2;
            n = n / (i + 2);
        }
    }
     
    return n > 4 ? n : maxPrime;
}
 
document.write(maxPrimeFactor(15));
document.write(maxPrimeFactor(25698751364526));
 
// This code is contributed by 8chkh7v1i3lrbhuvxyg3d9gbg9e097eodfhj7qbz.
</script>

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
        $n >>= 1;
    }
    // n must be odd at this point
    while ($n % 3 == 0) {
        $maxPrime = 3;
        $n=$n/3;
    }
    // now we have to iterate only for integers
    // who does not have prime factor 2 and 3
    for ($i = 3; $i <= sqrt($n); $i += 2)
    {
        while ($n % $i == 0)
        {
            $maxPrime = $i;
            $n = $n / $i;
        }
        while ($n % ($i+2) == 0) {
            $maxPrime = $i+2;
            $n = $n / ($i+2);
        }
    }
 
    // This condition is
    // to handle the case
    // when n is a prime
    // number greater than 4
    if ($n > 4)
        $maxPrime = $n;
 
    return $maxPrime;
}
 
    // Driver Code
    $n = 15;
    echo maxPrimeFactors($n), "\n";
 
    $n = 25698751364526;
    echo maxPrimeFactors($n), "\n";
 
 
?>

Output

5
328513

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

Method 2:

Follow the steps below for the implementation:

Initialize variables largest_prime to -1, i to 2, and n to the input integer.
Start a while loop that continues as long as i * i <= n. This loop will iterate through all possible factors of n.
In the while loop, start another while loop that continues as long as n % i == 0. This inner loop will divide n by i until n is no longer divisible by i.
In the inner loop, set largest_prime to i, and update n by dividing it by i.
At the end of the inner loop, increment i by 1.
After the outer loop, if n > 1, set largest_prime to n. This is because n could be a prime number larger than any of its factors.
Return largest_prime.

Below is the implementation of the above approach:

C++

// C++ code to find largest prime
// factor of number
#include <bits/stdc++.h>
using namespace std;
 
int maxPrimeFactors(long long n)
{
    /*
     * Find the largest prime factor of a positive integer
     * 'n'
     * @param n: positive integer (1 <= n <= 10^15)
     * @return: largest prime factor of n
     */
    int largest_prime = -1;
    int i = 2;
    while (i * i <= n) {
        while (n % i == 0) {
            largest_prime = i;
            n = n / i;
        }
        i = i + 1;
    }
    if (n > 1) {
        largest_prime = n;
    }
    return largest_prime;
}
 
int main()
{
    long long n = 15;
    cout << maxPrimeFactors(n) << endl;
 
    n = 25698751;
    cout << maxPrimeFactors(n) << endl;
 
    return 0;
}
 
// This code is contributed by Susobhan Akhuli

Java

/*package whatever //do not write package name here */
 
import java.io.*;
 
class GFG {
      static long maxPrimeFactors(long n)
    {
        /*
         * Find the largest prime factor of a positive integer
         * 'n'
         * @param n: positive integer (1 <= n <= 10^15)
         * @return: largest prime factor of n
         */
        long largest_prime = -1;
        long i = 2;
        while (i * i <= n) {
            while (n % i == 0) {
                largest_prime = i;
                n = n / i;
            }
            i = i + 1;
        }
        if (n > 1) {
            largest_prime = n;
        }
        return largest_prime;
    }
    public static void main (String[] args) {
        long n = 15;
        System.out.println(maxPrimeFactors(n));
 
        n = 25698751;
        System.out.println(maxPrimeFactors(n));
 
    }
}
//code contributed by shubhamrajput6156

Javascript

// JS code to find largest prime
// factor of number
 
function maxPrimeFactors(n)
{
    /*
     * Find the largest prime factor of a positive leteger
     * 'n'
     * @param n: positive leteger (1 <= n <= 10^15)
     * @return: largest prime factor of n
     */
    let largest_prime = -1;
    let i = 2;
    while (i * i <= n) {
        while (n % i == 0) {
            largest_prime = i;
            n = n / i;
        }
        i = i + 1;
    }
    if (n > 1) {
        largest_prime = n;
    }
    return largest_prime;
}
 
let n = 15;
document.write(maxPrimeFactors(n));
 
n = 25698751;
document.write(maxPrimeFactors(n));