# Count prime factors of N!

Count prime factors of N!

Given an integer N, the task is to count the number of prime factors of N!.

Examples:

Input: N = 5Output: 3Explanation: Factorial of 5 = 120. Prime factors of 120 are {2, 3, 5}. Therefore, the count is 3.

Input: N = 1Output: 0

Naive Approach: Follow the steps to solve the problem :Initialize a variable, say fac, to store the factorial of a number.

Initialize a variable, say count, to count the prime factors of N!.

Iterate over the range [2, fac] and if the number is not prime, increment count.

Print the count as the answer.

Below is the implementation of the above approach:

C++

#include

using namespace std;

int factorial(int f)

{

if (f == 0 || f == 1) {

return 1;

}

else {

return (f * factorial(f – 1));

}

}

bool isPrime(int element)

{

for (int i = 2;

i