Count digits in a factorial using Kamenetsky’s Formula

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
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