# Count digits in a factorial | Set 2

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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 above idea :

## C++

 `// A optimised program to find the  ` `// number of digits in a factorial ` `#include ` `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

 ` `

Output:

```1
363233781
8565705523
199
```

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : nitin mittal, Mithun Kumar

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