# Efficient program to print all prime factors of a given number

Given a number n, write an efficient function to print all prime factors of n. For example, if the input number is 12, then the output should be “2 2 3”. And if the input number is 315, then the output should be “3 3 5 7”.

### First Approach:

Following are the steps to find all prime factors.
1) While n is divisible by 2, print 2 and divide n by 2.
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 and continue.
3) If n is a prime number and is greater than 2, then n will not become 1 by the above two steps. So print n if it is greater than 2.

## C++

 // C++ program to print all prime factors  #include 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 <= sqrt(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 << " ";  }     /* Driver code */ int main()  {      int n = 315;      primeFactors(n);      return 0;  }     // This is code is contributed by rathbhupendra

## C

 // C Program to print all prime factors # include    // 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); }    /* Driver program to test above function */ int main() {     int n = 315;     primeFactors(n);     return 0; }    // This is code is improved by Susobhan Akhuli

## 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     public 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 <= Math.sqrt(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

 # 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            # Driver Program to test above function    n = 315 primeFactors(n)    # This code is contributed by Harshit Agrawal

## C#

 // C# Program to print all prime factors using System;    namespace prime { public class GFG {                             // A function to print all prime      // factors of a given number n     public 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 <= Math.Sqrt(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);     }            // Driver Code     public static void Main()     {         int n = 315;         primeFactors(n);     }    }  }    // This code is contributed by Sam007

## PHP

 2)         echo \$n," "; }        // Driver Code     \$n = 315;     primeFactors(\$n);    // This code is contributed by aj_36 ?>

## Javascript



Output

3 3 5 7

Time Complexity: O(sqrt(n))

In the worst case ( when either n or sqrt(n) is prime, for example: take n=11 or n=121 for both the cases for loop runs sqrt(n) times), the for loop runs for sqrt(n) times. The more number of times the while loop iterates on a number it reduces the original n, which also reduces the value of sqrt(n). Although the best case time complexity is O(log(n)), when the prime factors of n is only 2 and 3 or n is of the form (2^x*(3^y) where x>=0 and y>=0.

Auxiliary Space: O(1)

How does this work?
The steps 1 and 2 take care of composite numbers and step 3 takes care of prime numbers. To prove that the complete algorithm works, we need to prove that steps 1 and 2 actually take care of composite numbers. This is clear that step 1 takes care of even numbers. And after step 1, all remaining prime factors must be odd (difference of two prime factors must be at least 2), this explains why i is incremented by 2.

Now the main part is, the loop runs till the square root of n not till n. To prove that this optimization works, let us consider the following property of composite numbers.

Every composite number has at least one prime factor less than or equal to the square root of itself.
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 âˆšn, then a.b > âˆšn, * âˆšn, which contradicts the expression “a * b = n”.

In step 2 of the above algorithm, we run a loop and do the following in loop
a) Find the least prime factor i (must be less than âˆšn,)
b) Remove all occurrences i from n by repeatedly dividing n by i.
c) Repeat steps a and b for divided n and i = i + 2. The steps a and b are repeated till n becomes either 1 or a prime number.

Related Article :
Prime Factorization using Sieve O(log n) for multiple queries
Thanks to Vishwas Garg for suggesting the above algorithm. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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