# 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 ` `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

 `=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 :

```1
3
7
199
```

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

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.