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Count digits in a factorial | Set 2

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  • Difficulty Level : Medium
  • Last Updated : 24 Sep, 2022
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Given an integer n (can be very large), 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

Input : 50000000
Output : 363233781

Input : 1000000000
Output : 8565705523

 

We’ve already discussed the solution for small values of n in the Count digits in a factorial | Set 1. However that solution would not be able to handle cases where n >10^6 
So, can we improve our solution? 
Yes! we can. 
We can use Kamenetsky’s formula to find our answer! 
 

It approximates the number of digits in a factorial by :
f(x) =    log10(((n/e)^n) * sqrt(2*pi*n))

Thus, we can pretty easily use the property of logarithms to,
f(x) = n* log10((n/e)) + log10(2*pi*n)/2 

And that’s it! 
Our solution can handle very large inputs that can be accommodated in a 32-bit integer, 
and even beyond that! 

Below is the implementation of the above idea :

C++




// A optimised program to find the
// number of digits in a factorial
#include <bits/stdc++.h>
using namespace std;
 
// Returns the number of digits present
// in n! Since the result can be large
// long long is used as return type
long long findDigits(int n)
{
    // factorial of -ve number
    // doesn't exists
    if (n < 0)
        return 0;
 
    // base case
    if (n <= 1)
        return 1;
 
    // Use Kamenetsky formula to calculate
    // the number of digits
    double x = ((n * log10(n / M_E) +
                 log10(2 * M_PI * n) /
                 2.0));
 
    return floor(x) + 1;
}
 
// Driver Code
int main()
{
    cout << findDigits(1) << endl;
    cout << findDigits(50000000) << endl;
    cout << findDigits(1000000000) << endl;
    cout << findDigits(120) << endl;
    return 0;
}

Java




// An optimised java program to find the
// number of digits in a factorial
import java.io.*;
import java.util.*;
 
class GFG {
    public static double M_E = 2.71828182845904523536;
    public static double M_PI = 3.141592654;
 
     // Function returns the number of
     // digits present in n! since the
     // result can be large, long long
     // is used as return type
    static long findDigits(int n)
    {
        // factorial of -ve number doesn't exists
        if (n < 0)
            return 0;
 
        // base case
        if (n <= 1)
            return 1;
 
        // Use Kamenetsky formula to calculate
        // the number of digits
        double x = (n * Math.log10(n / M_E) +
                    Math.log10(2 * M_PI * n) /
                    2.0);
 
        return (long)Math.floor(x) + 1;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        System.out.println(findDigits(1));
        System.out.println(findDigits(50000000));
        System.out.println(findDigits(1000000000));
        System.out.println(findDigits(120));
    }
}
 
// This code is contributed by Pramod Kumar.

Python3




# A optimised Python3 program to find
# the number of digits in a factorial
import math
 
# Returns the number of digits present
# in n! Since the result can be large
# long long is used as return type
def findDigits(n):
     
    # factorial of -ve number
    # doesn't exists
    if (n < 0):
        return 0;
 
    # base case
    if (n <= 1):
        return 1;
 
    # Use Kamenetsky formula to
    # calculate the number of digits
    x = ((n * math.log10(n / math.e) +
              math.log10(2 * math.pi * n) /2.0));
 
    return math.floor(x) + 1;
 
# Driver Code
print(findDigits(1));
print(findDigits(50000000));
print(findDigits(1000000000));
print(findDigits(120));
     
# This code is contributed by mits

C#




// An optimised C# program to find the
// number of digits in a factorial.
using System;
 
class GFG {
    public static double M_E = 2.71828182845904523536;
    public static double M_PI = 3.141592654;
 
    // Function returns the number of
    // digits present in n! since the
    // result can be large, long long
    // is used as return type
    static long findDigits(int n)
    {
        // factorial of -ve number
        // doesn't exists
        if (n < 0)
            return 0;
 
        // base case
        if (n <= 1)
            return 1;
 
        // Use Kamenetsky formula to calculate
        // the number of digits
        double x = (n * Math.Log10(n / M_E) +
                    Math.Log10(2 * M_PI * n) /
                    2.0);
 
        return (long)Math.Floor(x) + 1;
    }
 
    // Driver Code
    public static void Main()
    {
        Console.WriteLine(findDigits(1));
        Console.WriteLine(findDigits(50000000));
        Console.WriteLine(findDigits(1000000000));
        Console.Write(findDigits(120));
    }
}
 
// This code is contributed by Nitin Mittal

PHP




<?php
// A optimised PHP program to find the
// number of digits in a factorial
 
// Returns the number of digits present
// in n! Since the result can be large
// long long is used as return type
function findDigits($n)
{
     
    // factorial of -ve number
    // doesn't exists
    if ($n < 0)
        return 0;
 
    // base case
    if ($n <= 1)
        return 1;
 
    // Use Kamenetsky formula to
    // calculate the number of digits
    $x = (($n * log10($n / M_E) +
                log10(2 * M_PI * $n) /
                2.0));
 
    return floor($x) + 1;
}
 
    // Driver Code
    echo findDigits(1)."\n" ;
    echo findDigits(50000000)."\n" ;
    echo findDigits(1000000000)."\n" ;
    echo findDigits(120) ;
     
// This code is contributed by nitin mittal
?>

Javascript




<script>
 
// A optimised Javascript program to find the
// number of digits in a factorial 
 
// Returns the number of digits present
// in n! Since the result can be large
// long long is used as return type
function findDigits(n)
{
    // factorial of -ve number
    // doesn't exists
    if (n < 0)
        return 0;
 
    // base case
    if (n <= 1)
        return 1;
 
    // Use Kamenetsky formula to calculate
    // the number of digits
    let x = ((n * Math.log10(n / Math.E) +
                Math.log10(2 * Math.PI * n) /
                2.0));
 
    return Math.floor(x) + 1;
}
 
// Driver Code
  
    document.write(findDigits(1) + "<br>");
    document.write(findDigits(50000000) + "<br>");
    document.write(findDigits(1000000000) + "<br>");
    document.write(findDigits(120) + "<br>");
 
// This code is contributed by Mayank Tyagi
 
</script>

Output

1
363233781
8565705523
199

Time complexity: O(logn)
Auxiliary space: O(1)

Method:  First finding the factorial of a number using factorial function then using while loop finding the number of digits present in the factorial number.
 

C++




// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
int factorial(int n)
{
    int fact = 1;
    for (int i = 1; i <= n; i++) {
        fact = fact * i;
    }
    return fact;
}
 
int main()
{
    int n = 10, c = 0;
   
    // finding factorial of a number
    int f = factorial(n);
   
    // counting the number of digits present
    // in the factoral number
    while (f != 0) {
        f /= 10;
        c += 1;
    }
    cout << c << endl;
}
 
// This code is contributed by phasing17

Java




// Java code for the above approach
import java.util.*;
 
class GFG
{
  static int factorial(int n)
  {
    int fact = 1;
    for (int i = 1; i <= n; i++) {
      fact = fact * i;
    }
    return fact;
  }
 
  public static void main(String[] args)
  {
    int n = 10, c = 0;
 
    // finding factorial of a number
    int f = factorial(n);
 
    // counting the number of digits present
    // in the factoral number
    while (f != 0) {
      f /= 10;
      c += 1;
    }
    System.out.println(c);
  }
}
 
// This code is contributed by phasing17

Python3




# Python code to count number of digits
# in a factorial of a number
 
# importing math module
from math import*
n=10;c=0
# finding factorial of a number
f=factorial(n)
# counting the number of digits present
# in the factoral number
while(f!=0):
  f//=10
  c+=1
print(c)
 
 
# this code is contributed by gangarajula laxmi

C#




// C# code for the above approach
using System;
using System.Collections.Generic;
 
class GFG
{
  static int factorial(int n)
  {
    int fact = 1;
    for (int i = 1; i <= n; i++) {
      fact = fact * i;
    }
    return fact;
  }
 
  public static void Main(string[] args)
  {
    int n = 10, c = 0;
 
    // finding factorial of a number
    int f = factorial(n);
 
    // counting the number of digits present
    // in the factoral number
    while (f != 0) {
      f /= 10;
      c += 1;
    }
    Console.WriteLine(c);
  }
}
 
// This code is contributed by phasing17

Javascript




<script>
       // JavaScript code for the above approach
       function factorial(n) {
           let fact = 1;
           for (let i = 1; i <= n; i++) {
               fact = fact * i;
           }
           return fact;
       }
       let n = 10; c = 0
       // finding factorial of a number
       let f = factorial(n)
       // counting the number of digits present
       // in the factoral number
       while (f != 0) {
           f = Math.floor(f / 10)
           c += 1
       }
       document.write(c);
   // This code is contributed by Potta Lokesh
   </script>

Output

7

Time complexity: O(n) because factorial function is using a for loop 
Auxiliary space: O(1) as it is using constant space for variables

References : oeis.org 
This article is contributed by Ashutosh Kumar .If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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