Count digits in a factorial | Set 1

Given an integer n, find the number of digits that appear in its factorial, where factorial is defined as, factorial(n) = 1*2*3*4……..*n and factorial(0) = 1

Examples :

Input :  n = 1
Output : 1
1! = 1 , hence number of digits is 1

Input :  5
Output : 3
5! = 120, i.e., 3 digits

Input : 10
Output : 7
10! = 3628800, i.e., 7 digits

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive solution would be to calculate the n! first and then calculate the number of digits present in it. However as the value for n! can be very large, it would become cumbersome to store them in a variable (Unless you’re working in python!) .
A better solution would be to use the useful property of logarithms to calculate the required answer.

We know,
log(a*b) = log(a) + log(b)

Therefore
log( n! ) = log(1*2*3....... * n) 
          = log(1) + log(2) + ........ +log(n)

Now, observe that the floor value of log base 
10 increased by 1, of any number, gives the
number of digits present in that number.

Hence, output would be : floor(log(n!)) + 1.

Below is a the implementation of the same.

C++

// A C++ program to find the number of digits in 
// a factorial 
#include <bits/stdc++.h> 
using namespace std; 
  
// This function receives an integer n, and returns 
// the number of digits present in n! 
int findDigits(int n) 
{ 
    // factorial exists only for n>=0 
    if (n < 0) 
        return 0; 
  
    // base case 
    if (n <= 1) 
        return 1; 
  
    // else iterate through n and calculate the 
    // value 
    double digits = 0; 
    for (int i=2; i<=n; i++) 
        digits += log10(i); 
  
    return floor(digits) + 1; 
} 
  
// Driver code  
int main() 
{ 
    cout << findDigits(1) << endl; 
    cout << findDigits(5) << endl; 
    cout << findDigits(10) << endl; 
    cout << findDigits(120) << endl; 
    return 0; 
} 

Java

// Java program to find the number  
// of digits in a factorial 
  
import java.io.*; 
import java.util.*; 
  
class GFG  
{ 
    // returns the number of digits  
    // present in n! 
    static int findDigits(int n) 
    { 
        // factorial exists only for n>=0 
        if (n < 0) 
            return 0; 
   
        // base case 
        if (n <= 1) 
            return 1; 
   
        // else iterate through n and calculate the 
        // value 
        double digits = 0; 
        for (int i=2; i<=n; i++) 
            digits += Math.log10(i); 
   
        return (int)(Math.floor(digits)) + 1; 
    } 
      
    // Driver code 
    public static void main (String[] args)  
    { 
        System.out.println(findDigits(1)); 
        System.out.println(findDigits(5)); 
        System.out.println(findDigits(10)); 
        System.out.println(findDigits(120)); 
    } 
} 
  
// This code is contributed by Pramod Kumar 

Python3

# Python3 program to find the  
# number of digits in a factorial 
import math 
  
# This function receives an integer  
# n, and returns the number of 
# digits present in n! 
  
def findDigits(n): 
      
    # factorial exists only for n>=0 
    if (n < 0): 
        return 0; 
  
    # base case 
    if (n <= 1): 
        return 1; 
  
    # else iterate through n and  
    # calculate the value 
    digits = 0; 
    for i in range(2, n + 1): 
        digits += math.log10(i); 
  
    return math.floor(digits) + 1; 
  
# Driver code  
print(findDigits(1)); 
print(findDigits(5)); 
print(findDigits(10)); 
print(findDigits(120)); 
  
# This code is contributed by mits 

C#

// A C++ program to find the number 
// of digits in a factorial 
using System; 
  
class GFG { 
      
    // This function receives an integer 
    // n, and returns the number of  
    // digits present in n! 
    static int findDigits(int n) 
    { 
          
        // factorial exists only for n>=0 
        if (n < 0) 
            return 0; 
      
        // base case 
        if (n <= 1) 
            return 1; 
      
        // else iterate through n and  
        // calculate the value 
        double digits = 0; 
        for (int i = 2; i <= n; i++) 
            digits += Math.Log10(i); 
      
        return (int)Math.Floor(digits) + 1; 
    } 
      
    // Driver code  
    public static void Main() 
    { 
        Console.Write(findDigits(1) + "\n"); 
        Console.Write(findDigits(5) + "\n"); 
        Console.Write(findDigits(10) + "\n"); 
        Console.Write(findDigits(120) + "\n"); 
    } 
} 
  
// This code is contributed by 
// Smitha Dinesh Semwal 

PHP

<?php 
// PHP program to find  
// the number of digits  
// in a factorial 
  
// This function receives  
// an integer n, and returns 
// the number of digits present in n! 
  
function findDigits($n) 
{ 
    // factorial exists only for n>=0 
    if ($n < 0) 
        return 0; 
  
    // base case 
    if ($n <= 1) 
        return 1; 
  
    // else iterate through n and  
    // calculate the value 
    $digits = 0; 
    for ($i = 2; $i <= $n; $i++) 
        $digits += log10($i); 
  
    return floor($digits) + 1; 
} 
  
// Driver code  
echo findDigits(1), "\n"; 
echo findDigits(5), "\n"; 
echo findDigits(10), "\n"; 
echo findDigits(120), "\n"; 
  
// This code is contributed by Ajit. 
?> 

Output :

In the next set, we’d see how to further optimize our approach and reduce the time complexity for the same program.

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