Print all prime factors of a given number

Try it on GfG Practice

Given a number n, the task is to find all prime factors of n.

Examples:

Input: n = 24
Output: 2 2 2 3
Explanation: The prime factorization of 24 is 2³×3.

Input: n = 13195
Output: 5 7 13 29
Explanation: The prime factorization of 13195 is 5×7×13×29.

Approach:

Every composite number has at least one prime factor less than or equal to its square root. This property can be proved using a counter statement -

• Let a and b be two factors of n such that a*b = n.
• If both are greater than sqrt(n), a*b > sqrt(n)*sqrt(n), which contradicts the expression a * b = n

Following are the steps to find all prime factors:

• While n is divisible by 2, print 2 and divide n by 2.
• After step 1, n must be odd. Now start a loop from i = 3 to the square root of n. While i divides n, print i, and divide n by i. After i fails to divide n, increment i by 2 because next prime factor will also be odd since we have already taken care of all the 2 and continue.
• If n greater than 2, then n is the last prime factor.

// C++ program to print all prime factors 
#include <bits/stdc++.h>
using namespace std;

// A function to print all prime 
// factors of a given number n 
void primeFactors(int n) {
  
    // Print the number of 2s that divide n 
    while (n % 2 == 0) {
      
        cout << 2 << " "; 
        n = n/2; 
    } 

    // n must be odd at this point. So we can skip 
    // one element (Note i = i +2) 
    for (int i = 3; i*i <= n; i = i + 2) {
      
        // While i divides n, print i and divide n 
        while (n % i == 0) {
          
            cout << i << " "; 
            n = n/i; 
        } 
    } 

    // This condition is to handle the case when n 
    // is a prime number greater than 2 
    if (n > 2) 
        cout << n << " "; 
} 


int main() {
  
    int n = 315; 
    primeFactors(n); 
    return 0; 
} 

// C Program to print all prime factors
# include <stdio.h>

// A function to print all prime factors of a given number n
void primeFactors(int n) {
  
    // Print the number of 2s that divide n
    while (n%2 == 0) {
      
        printf("%d ", 2);
        n = n/2;
    }

    // n must be odd at this point.  So we can skip 
    // one element (Note i = i +2)
    for (int i = 3; i*i <= n; i = i+2) {
      
        // While i divides n, print i and divide n
        while (n%i == 0){
          
            printf("%d ", i);
            n = n/i;
        }
    }

    // This condition is to handle the case when n 
    // is a prime number greater than 2
    if (n > 2)
        printf ("%d ", n);
}


int main() {
  
    int n = 315;
    primeFactors(n);
    return 0;
}

// Java Program to print all prime factors
import java.io.*;
import java.lang.Math;

class GfG {

    // A function to print all prime factors
    // of a given number n
    static void primeFactors(int n) {

        // Print the number of 2s that divide n
        while (n % 2 == 0) {

            System.out.print(2 + " ");
            n /= 2;
        }

        // n must be odd at this point.  So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i * i <= n; i += 2) {

            // While i divides n, print i and divide n
            while (n % i == 0) {

                System.out.print(i + " ");
                n /= i;
            }
        }

        // This condition is to handle the case when
        // n is a prime number greater than 2
        if (n > 2)
            System.out.print(n);
    }

    public static void main(String[] args) {

        int n = 315;
        primeFactors(n);
    }
}

# Python program to print prime factors

import math

# A function to print all prime factors of 
# a given number n
def primeFactors(n):
    
    # Print the number of two's that divide n
    while n % 2 == 0:
        print 2,
        n = n / 2
        
    # n must be odd at this point
    # so a skip of 2 ( i = i + 2) can be used
    for i in range(3,int(math.sqrt(n))+1,2):
        
        # while i divides n , print i and divide n
        while n % i== 0:
            print i,
            n = n / i
            
    # Condition if n is a prime
    # number greater than 2
    if n > 2:
        print n

n = 315
primeFactors(n)

// C# Program to print all prime factors
using System;

class GfG {

    // A function to print all prime
    // factors of a given number n
    static void primeFactors(int n) {

        // Print the number of 2s that divide n
        while (n % 2 == 0) {

            Console.Write(2 + " ");
            n /= 2;
        }

        // n must be odd at this point. So we can
        // skip one element (Note i = i +2)
        for (int i = 3; i * i <= n; i += 2) {

            // While i divides n, print i and divide n
            while (n % i == 0) {

                Console.Write(i + " ");
                n /= i;
            }
        }

        // This condition is to handle the case when
        // n is a prime number greater than 2
        if (n > 2)
            Console.Write(n);
    }

    static void Main() {

        int n = 315;
        primeFactors(n);
    }
}

// JavaScript program to print all prime factors

// A function to print all prime
// factors of a given number n
function primeFactors(n) {

    // Print the number of 2s that divide n
    while (n % 2 == 0) {

        console.log(2 + " ");
        n = Math.floor(n / 2);
    }

    // n must be odd at this point.
    // So we can skip one element
    // (Note i = i +2)

    for (let i = 3; i <= Math.floor(Math.sqrt(n));
         i = i + 2) {

        // While i divides n, print i
        // and divide n
        while (n % i == 0) {

            process.stdout.write(i + " ");
            n = Math.floor(n / i);
        }
    }

    // This condition is to handle the
    // case when n is a prime number
    // greater than 2
    if (n > 2)
        process.stdout.write(n + " ");
}

let n = 315;
primeFactors(n);

3 3 5 7 

Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)

Related article: 

• Prime Factorization using Sieve O(log n) for multiple queries