Count prime factors of N!

Namachila November 19, 2025

0 0 9 minutes read

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

Examples:

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

Input: N = 1
Output: 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++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate
// factorial of a number
int factorial(int f)
{
    // Base Case
    if (f == 0 || f == 1) {
        return 1;
    }
    else {
 
        // Recursive call
        return (f * factorial(f - 1));
    }
}
 
// Function to check if a
// number is prime or not
bool isPrime(int element)
{
    for (int i = 2;
         i <= sqrt(element); i++) {
        if (element % i == 0) {
 
            // Not prime
            return false;
        }
    }
 
    // Is prime
    return true;
}
 
// Function to count the number
// of prime factors of N!
int countPrimeFactors(int N)
{
    // Stores factorial of N
    int fac = factorial(N);
 
    // Stores the count of
    // prime factors
    int count = 0;
 
    // Iterate over the range [2, fac]
    for (int i = 2; i <= fac; i++) {
 
        // If not prime
        if (fac % i == 0 && isPrime(i)) {
 
            // Increment count
            count++;
        }
    }
 
    // Print the count
    cout << count;
}
 
// Driver Code
int main()
{
    // Given value of N
    int N = 5;
 
    // Function call to count the
    // number of prime factors of N
    countPrimeFactors(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG
{
 
  // Function to calculate
  // factorial of a number
  static int factorial(int f)
  {
    // Base Case
    if (f == 0 || f == 1) {
      return 1;
    }
    else {
 
      // Recursive call
      return (f * factorial(f - 1));
    }
  }
 
  // Function to check if a
  // number is prime or not
  static boolean isPrime(int element)
  {
    for (int i = 2;
         i <= (int)Math.sqrt(element); i++) {
      if (element % i == 0) {
 
        // Not prime
        return false;
      }
    }
 
    // Is prime
    return true;
  }
 
  // Function to count the number
  // of prime factors of N!
  static void countPrimeFactors(int N)
  {
    // Stores factorial of N
    int fac = factorial(N);
 
    // Stores the count of
    // prime factors
    int count = 0;
 
    // Iterate over the range [2, fac]
    for (int i = 2; i <= fac; i++) {
 
      // If not prime
      if ((fac % i == 0 && isPrime(i))) {
 
        // Increment count
        count++;
      }
    }
 
    // Print the count
    System.out.println(count);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
 
    // Given value of N
    int N = 5;
 
    // Function call to count the
    // number of prime factors of N
    countPrimeFactors(N);
  }
}
 
// This code is contributed by sanjoy_62.

Python3

# Python program for the above approach
from math import sqrt
 
# Function to calculate
# factorial of a number
def factorial(f):
     
    # Base Case
    if (f == 0 or f == 1):
        return 1
    else:
       
        # Recursive call
        return (f * factorial(f - 1))
         
# Function to check if a
# number is prime or not
def isPrime(element):
     
    for i in range(2,int(sqrt(element))+1):
        if (element % i == 0):
           
            # Not prime
            return False
     
    # Is prime
    return True
 
# Function to count the number
# of prime factors of N!
def countPrimeFactors(N):
     
    # Stores factorial of N
    fac = factorial(N)
     
    # Stores the count of
    # prime factors
    count = 0
     
    # Iterate over the range [2, fac]
    for i in range(2, fac + 1):
         
        # If not prime
        if (fac % i == 0 and isPrime(i)):
             
            # Increment count
            count += 1
             
    # Print the count
    print(count)
 
# Driver Code
# Given value of N
N = 5
 
# Function call to count the
# number of prime factors of N
countPrimeFactors(N)
 
# This code is contributed by shubhamsingh10

C#

// C# program for the above approach
using System;
 
class GFG
{
 
  // Function to calculate
  // factorial of a number
  static int factorial(int f)
  {
     
    // Base Case
    if (f == 0 || f == 1) {
      return 1;
    }
    else {
  
      // Recursive call
      return (f * factorial(f - 1));
    }
  }
  
  // Function to check if a
  // number is prime or not
  static bool isPrime(int element)
  {
    for (int i = 2;
         i <= (int)Math.Sqrt(element); i++) {
      if (element % i == 0) {
  
        // Not prime
        return false;
      }
    }
  
    // Is prime
    return true;
  }
  
  // Function to count the number
  // of prime factors of N!
  static void countPrimeFactors(int N)
  {
     
    // Stores factorial of N
    int fac = factorial(N);
  
    // Stores the count of
    // prime factors
    int count = 0;
  
    // Iterate over the range [2, fac]
    for (int i = 2; i <= fac; i++) {
  
      // If not prime
      if ((fac % i == 0 && isPrime(i))) {
  
        // Increment count
        count++;
      }
    }
  
    // Print the count
    Console.Write(count);
  }
 
 
// Driver Code
public static void Main()
{
   
    // Given value of N
    int N = 5;
  
    // Function call to count the
    // number of prime factors of N
    countPrimeFactors(N);
}
}
 
// This code is contributed by code_hunt.

Javascript

<script>
// Javascript program for the above approach
 
// Function to calculate
// factorial of a number
function factorial(f){
     
    // Base Case
    if (f == 0 || f == 1){
        return 1
    }
    else{
         // Recursive call
         return (f * factorial(f - 1))
        }
    }
         
// Function to check if a
// number is prime or not
function isPrime(element){
     
    for (let i = 2;  i < Math.floor(Math.sqrt(element)+1); i++){
        if (element % i == 0){
            // Not prime
            return false
        }
    }
    // Is prime
    return true
}
 
// Function to count the number
// of prime factors of N!
function countPrimeFactors(N){
     
    // Stores factorial of N
    let fac = factorial(N)
     
    // Stores the count of
    // prime factors
    let count = 0
     
    // Iterate over the range [2, fac]
    for(let i = 2;  i < fac + 1; i++){
         
        // If not prime
        if (fac % i == 0 && isPrime(i)){    
            // Increment count
            count += 1
        }
    }
             
    // Print the count
    document.write(count)
}
 
// Driver Code
// Given value of N
let N = 5
 
// Function call to count the
// number of prime factors of N
countPrimeFactors(N)
// This code is contributed by _saurabh_jaiswal
</script>

Output:

Time Complexity: O(N! * sqrt(N))
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use Sieve Of Eratosthenes. Follow the steps below to solve the problem:

Initialize a variable, say count, to store the count of prime factors of N!.
Initialize a boolean array, say prime[] to check if a number is prime or not.
Perform Sieve of Eratosthenes and populate count at each iteration, if found prime.
Print the value of count as the answer.

Below is the implementation of the above approach:

C++

// C++ approach for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the
// prime factors of N!
int countPrimeFactors(int N)
{
    // Stores the count of
    // prime factors
    int count = 0;
 
    // Stores whether a number
    // is prime or not
    bool prime[N + 1];
 
    // Mark all as true initially
    memset(prime, true, sizeof(prime));
 
    // Sieve of Eratosthenes
    for (int p = 2; p * p <= N; p++) {
 
        // If prime[p] is not changed,
        // then it is a prime
        if (prime[p] == true) {
 
            // Update all subsequent multiples
            for (int i = p * p; i <= N; i += p)
                prime[i] = false;
        }
    }
 
    // Traverse in the range [2, N]
    for (int p = 2; p <= N; p++) {
 
        // If prime
        if (prime[p]) {
 
            // Increment the count
            count++;
        }
    }
 
    // Print the count
    cout << count;
}
 
// Driver Code
int main()
{
    // Given  value of N
    int N = 5;
 
    // Function call to count
    // the prime factors of N!
    countPrimeFactors(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG 
{
 
  // Function to count the
  // prime factors of N!
  static void countPrimeFactors(int N)
  {
     
    // Stores the count of
    // prime factors
    int count = 0;
 
    // Stores whether a number
    // is prime or not
    boolean[] prime = new boolean[N + 1];
 
    // Mark all as true initially
    Arrays.fill(prime, true);
 
    // Sieve of Eratosthenes
    for (int p = 2; p * p <= N; p++) {
 
      // If prime[p] is not changed,
      // then it is a prime
      if (prime[p] == true) {
 
        // Update all subsequent multiples
        for (int i = p * p; i <= N; i += p)
          prime[i] = false;
      }
    }
 
    // Traverse in the range [2, N]
    for (int p = 2; p <= N; p++) {
 
      // If prime
      if (prime[p] != false) {
 
        // Increment the count
        count++;
      }
    }
 
    // Print the count
    System.out.print(count);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
     
    // Given  value of N
    int N = 5;
 
    // Function call to count
    // the prime factors of N!
    countPrimeFactors(N);
  }
}
 
// This code is contributed by susmitakundugoaldanga.

Python3

# Python3 approach for the above approach
 
# Function to count the
# prime factors of N!
def countPrimeFactors(N):
     
    # Stores the count of
    # prime factors
    count = 0
 
    # Stores whether a number
    # is prime or not
    prime = [1] * (N + 1)
 
    # Sieve of Eratosthenes
    for p in range(2, N + 1):
        if p * p > N:
            break
         
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p]):
 
            # Update all subsequent multiples
            for i in range(p * p, N + 1, p):
                prime[i] = 0
 
    # Traverse in the range [2, N]
    for p in range(2, N + 1):
         
        # If prime
        if (prime[p]):
 
            # Increment the count
            count += 1
 
    # Print the count
    print (count)
 
# Driver Code
if __name__ == '__main__':
     
    # Given  value of N
    N = 5
 
    # Function call to count
    # the prime factors of N!
    countPrimeFactors(N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
public class GFG 
{
 
  // Function to count the
  // prime factors of N!
  static void countPrimeFactors(int N)
  {
     
    // Stores the count of
    // prime factors
    int count = 0;
 
    // Stores whether a number
    // is prime or not
    bool[] prime = new bool[N + 1];
 
    // Mark all as true initially
    for (int i = 0; i < prime.Length; i++)
        prime[i] = true;
 
    // Sieve of Eratosthenes
    for (int p = 2; p * p <= N; p++) {
 
      // If prime[p] is not changed,
      // then it is a prime
      if (prime[p] == true) {
 
        // Update all subsequent multiples
        for (int i = p * p; i <= N; i += p)
          prime[i] = false;
      }
    }
 
    // Traverse in the range [2, N]
    for (int p = 2; p <= N; p++) {
 
      // If prime
      if (prime[p] != false) {
 
        // Increment the count
        count++;
      }
    }
 
    // Print the count
    Console.Write(count);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
     
    // Given  value of N
    int N = 5;
 
    // Function call to count
    // the prime factors of N!
    countPrimeFactors(N);
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

<script>
 
// JavaScript program for the above approach
 
     // Function to count the
    // prime factors of N!
    function countPrimeFactors( N) {
 
        // Stores the count of
        // prime factors
        var count = 0;
 
        // Stores whether a number
        // is prime or not
        var prime = Array(N + 1).fill(true);
 
     
 
        // Sieve of Eratosthenes
        for (var p = 2; p * p <= N; p++) {
 
            // If prime[p] is not changed,
            // then it is a prime
            if (prime[p] == true) {
 
                // Update all subsequent multiples
                for (var i = p * p; i <= N; i += p)
                    prime[i] = false;
            }
        }
 
        // Traverse in the range [2, N]
        for (var p = 2; p <= N; p++) {
 
            // If prime
            if (prime[p] != false) {
 
                // Increment the count
                count++;
            }
        }
 
        // Print the count
        document.write(count);
    }
 
    // Driver Code
     
 
        // Given value of N
        var N = 5;
 
        // Function call to count
        // the prime factors of N!
        countPrimeFactors(N);
 
// This code is contributed by Amit Katiyar
 
</script>

Output:

Time Complexity: O(N * log(logN))
Auxiliary Space: O(N)

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