Find largest prime factor of a number (original) (raw)

Last Updated : 17 Feb, 2025

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

**Input: 6
**Output: 3
**Explanation Prime factor are 2 and 3. Largest of them is **3.

**Input: 15
**Output: 5
**Explanation: Prime factors are 3 and 5. The largest of them is **5.

**Input: 28
**Output: 7
**Explanation: Prime factors are 2 and 7. The largest of them is **7.

Try It Yourself

[Naive Approach] Basic Trial Division

The method starts by removing all factors of 2, as it is the only even prime.

Once 2 is completely removed, odd numbers are checked starting from 3.

Each odd number is tested for divisibility, and the number is divided repeatedly until the factor is fully removed.

This process continues for all odd numbers up to √n.

If a number greater than 2 remains after all divisions, it is a prime number and the largest prime factor.

C++ `

// C++ code to find largest prime // factor of number

#include using namespace std;

int largestPrimeFactor(int n) {

int largestPrime = -1;

// Check for factors of 2
while (n % 2 == 0) {
    largestPrime = 2;
    n /= 2;
}

// Check for odd factors starting from 3
for (int i = 3; i * i <= n; i += 2) {
    while (n % i == 0)  {
        largestPrime = i;
        n /= i;
    }
}

// If n is still greater than 2, it is
// a prime number
if (n > 2) {
    largestPrime = n;
}

return largestPrime;

}

int main() { int n = 15; int res = largestPrimeFactor(n); cout << res << endl;

return 0;

}

Java

// Java code to find largest prime // factor of number

class GfG {

static int largestPrimeFactor(int n) {

    int largestPrime = -1;

    // Check for factors of 2
    while (n % 2 == 0) {
        largestPrime = 2;
        n /= 2;
    }

    // Check for odd factors starting from 3
    for (int i = 3; i * i <= n; i += 2) {
        while (n % i == 0) {
            largestPrime = i;
            n /= i;
        }
    }

    // If n is still greater than 2, it is
    // a prime number
    if (n > 2) {
        largestPrime = n;
    }

    return largestPrime;
}

public static void main(String[] args) {
    int n = 15;
    int res = largestPrimeFactor(n);
    System.out.println(res);
}

}

Python

Python code to find largest prime

factor of number

def largestPrimeFactor(n): largestPrime = -1

# Check for factors of 2
while n % 2 == 0:
    largestPrime = 2
    n //= 2

# Check for odd factors starting from 3
i = 3
while i * i <= n:
    while n % i == 0:
        largestPrime = i
        n //= i
    i += 2

# If n is still greater than 2, it is
# a prime number
if n > 2:
    largestPrime = n

return largestPrime

if name == "main": n = 15 res = largestPrimeFactor(n) print(res)

C#

// C# code to find largest prime // factor of number

using System;

class GfG {

static int largestPrimeFactor(int n) {
    int largestPrime = -1;

    // Check for factors of 2
    while (n % 2 == 0) {
        largestPrime = 2;
        n /= 2;
    }

    // Check for odd factors starting from 3
    for (int i = 3; i * i <= n; i += 2) {
        while (n % i == 0) {
            largestPrime = i;
            n /= i;
        }
    }

    // If n is still greater than 2, it is
    // a prime number
    if (n > 2) {
        largestPrime = n;
    }

    return largestPrime;
}

static void Main(string[] args) {
    int n = 15;
    int res = largestPrimeFactor(n);
    Console.WriteLine(res);
}

}

JavaScript

// JavaScript code to find largest prime // factor of number

function largestPrimeFactor(n) { let largestPrime = -1;

// Check for factors of 2
while (n % 2 === 0) {
    largestPrime = 2;
    n /= 2;
}

// Check for odd factors starting from 3
for (let i = 3; i * i <= n; i += 2) {
    while (n % i === 0) {
        largestPrime = i;
        n /= i;
    }
}

// If n is still greater than 2, it is a prime number
if (n > 2) {
    largestPrime = n;
}

return largestPrime;

}

let n = 15; let res = largestPrimeFactor(n); console.log(res);

**Time complexity: O(sqrt(n)).
**Auxiliary space: O(1)

[Expected Approach] Optimized Trial Division

The method first removes all factors of 2 and 3 to simplify the number.

After eliminating these smallest primes, further factorization follows a structured approach.

Instead of checking all odd numbers, only numbers of the form 6k ± 1 are tested.

This works because all prime numbers greater than 3 follow this pattern.

By skipping unnecessary checks, the approach reduces iterations while efficiently finding the largest prime factor.

C++ `

// C++ code to find largest prime // factor of number

#include using namespace std;

int largestPrimeFactor(int n) {

// Initialize the maximum prime factor variable
int maxPrime = -1;

// Check for factors of 2
while (n % 2 == 0) {
    maxPrime = 2;
    n >>= 1;  
}

// Check for factors of 3
while (n % 3 == 0) {
    maxPrime = 3;
    n = n / 3;
}

// Check for odd factors starting from 5 and

// incrementing by 6 (i and i+2) 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); } }

// If n is still greater than 4, it is a // prime number if (n > 4) maxPrime = n;

return maxPrime;

}

int main() { int n = 15; int res = largestPrimeFactor(n); cout << res << endl; return 0; }

C

// C code to find largest prime // factor of number

#include <stdio.h>

int largestPrimeFactor(int n) {

// Initialize the maximum prime factor variable
int maxPrime = -1;

// Check for factors of 2
while (n % 2 == 0) {
    maxPrime = 2;
    n /= 2;   
}

// Check for factors of 3
while (n % 3 == 0) {
    maxPrime = 3;
    n = n / 3;
}

// Check for odd factors starting from 5 and 
// incrementing by 6 (i and i+2)
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);
    }
}

// If n is still greater than 4, it is a prime

// number if (n > 4) maxPrime = n;

return maxPrime;

}

int main() { int n = 15; int res = largestPrimeFactor(n); printf("%d\n", res);
return 0; }

Java

// Java code to find largest prime // factor of number

class GfG {

static int largestPrimeFactor(int n) {

    // Initialize the maximum prime factor variable
    int maxPrime = -1;

    // Check for factors of 2
    while (n % 2 == 0) {
        maxPrime = 2;
        n /= 2;
    }

    // Check for factors of 3
    while (n % 3 == 0) {
        maxPrime = 3;
        n = n / 3;
    }

    // Check for odd factors starting from 5 and
    // incrementing by 6 (i and i+2)
    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);
        }
    }

    // If n is still greater than 4, it is a prime
    // number
    if (n > 4) {
        maxPrime = n;
    }

    return maxPrime;
}

public static void main(String[] args) {
    int n = 15;
    int res = largestPrimeFactor(n);
    System.out.println(res);
}

}

Python

Python code to find largest prime

factor of number

def largestPrimeFactor(n):

# Initialize the maximum prime factor variable
maxPrime = -1

# Check for factors of 2
while n % 2 == 0:
    maxPrime = 2
    n //= 2

# Check for factors of 3
while n % 3 == 0:
    maxPrime = 3
    n //= 3

# Check for odd factors starting from 5 and
# incrementing by 6 (i and i+2)
i = 5
while i * i <= n:
    while n % i == 0:
        maxPrime = i
        n //= i
    while n % (i + 2) == 0:
        maxPrime = i + 2
        n //= (i + 2)
    i += 6

# If n is still greater than 4, it is a prime
# number
if n > 4:
    maxPrime = n

return maxPrime

n = 15 res = largestPrimeFactor(n) print(res)

C#

// C# code to find largest prime // factor of number

using System;

class GfG {

static int largestPrimeFactor(int n) {

    // Initialize the maximum prime factor variable
    int maxPrime = -1;

    // Check for factors of 2
    while (n % 2 == 0) {
        maxPrime = 2;
        n /= 2;
    }

    // Check for factors of 3
    while (n % 3 == 0) {
        maxPrime = 3;
        n /= 3;
    }

    // Check for odd factors starting from 5 and
    // incrementing by 6 (i and i+2)
    for (int i = 5; i * i <= n; i += 6) {
        while (n % i == 0) {
            maxPrime = i;
            n /= i;
        }
        while (n % (i + 2) == 0) {
            maxPrime = i + 2;
            n /= (i + 2);
        }
    }

    // If n is still greater than 4, it is a prime
    // number
    if (n > 4) {
        maxPrime = n;
    }

    return maxPrime;
}

static void Main() {
    int n = 15;
    int res = largestPrimeFactor(n);
    Console.WriteLine(res);
}

}

JavaScript

// JavaScript code to find largest prime // factor of number

function largestPrimeFactor(n) {

// Initialize the maximum prime factor variable
let maxPrime = -1;

// Check for factors of 2
while (n % 2 === 0) {
    maxPrime = 2;
    n /= 2;
}

// Check for factors of 3
while (n % 3 === 0) {
    maxPrime = 3;
    n /= 3;
}

// Check for odd factors starting from 5 and 
// incrementing by 6 (i and i + 2)
for (let i = 5; i * i <= n; i += 6) {
    while (n % i === 0) {
        maxPrime = i;
        n /= i;
    }
    while (n % (i + 2) === 0) {
        maxPrime = i + 2;
        n /= (i + 2);
    }
}

// If n is still greater than 4, it is a prime number
if (n > 4) {
    maxPrime = n;
}

return maxPrime;

}

let n = 15; let res = largestPrimeFactor(n); console.log(res);

**Time complexity: O(sqrt(n))
**Auxiliary space: O(1)