# Number of digits in N factorial to the power N

Given a positive integer N, we have to find the total number of digits in the factorial of N raised to the power N, i.e, Examples:

```Input: 4
Output: 6
Explanations: = = 331776.
Total number of digits in 331776 is 6.

Input: 5
Output: 11
Explanations: = = 24883200000
Total number of digits in 24883200000 is 11.

Input: 2
Output: 1

Input: 1000
Output: 2567605```

The idea of the solution is described below.

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

Brute force method: By brute force, we can simply compute N! in O(N) time and then multiply it n times but that will be a very slow method and would exceed both time and space because would be a huge number.

Efficient method:
Let’s look at it more closely. We can break (N!)^N into simpler terms which are easy to compute. By taking common logarithm,

we get and we know, Therefore we can reduce further.   Now we can compute the answer easily in O(N) time and without space limit exceeding.

So why is a valid answer to the problem?
We know that the total number of digits in a number N whose base is 10 can be easily found by taking ceil(log10 (N)). That is exactly done in the approach described above.

The code implementation of the above idea is given below.

## C++

 `// CPP program to find count of digits in N  ` `// factorial raised to N ` `#include ` `using` `namespace` `std; ` ` `  `int` `countDigits(``int` `n) ` `{ ` `    ``// we take sum of logarithms as explained ` `    ``// in the approach ` `    ``double` `ans = 0; ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `        ``ans += ``log10``(i); ` ` `  `    ``// multiply the result with n ` `    ``ans = ans * n; ` `    ``return` `1 + ``floor``(ans); ` `} ` ` `  `int` `main() ` `{ ` `    ``int` `n = 4; ` `    ``cout << countDigits(n) << ``"\n"``; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find ` `// count of digits in N  ` `// factorial raised to N ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `static` `int` `countDigits(``int` `n) ` `{ ` `    ``// we take sum of logarithms  ` `    ``// as explained in the approach ` `    ``double` `ans = ``0``; ` `    ``for` `(``int` `i = ``1``; i <= n; i++) ` `        ``ans += Math.log10(i); ` ` `  `    ``// multiply the ` `    ``// result with n ` `    ``ans = ans * n; ` `    ``return` `1` `+ (``int``)Math.floor(ans); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args)  ` `{ ` `    ``int` `n = ``4``; ` `    ``System.out.println( ` `               ``countDigits(n) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed ` `// by anuj_67. `

## Python3

 `# python3 program to find count of digits in N  ` `# factorial raised to N ` ` `  `import` `math ` `  `  `def` `countDigits( n): ` ` `  `    ``# we take sum of logarithms as explained ` `    ``# in the approach ` `    ``ans ``=` `0` `    ``for` `i ``in` `range` `(``1``,n``+``1``): ` `        ``ans ``+``=` `math.log10(i) ` `  `  `    ``#multiply the result with n ` `    ``ans ``=` `ans ``*` `n ` `    ``return` `1` `+` `math.floor(ans) ` ` `  `  `  `if` `__name__ ``=``=` `"__main__"``: ` ` `  `    ``n ``=` `4` `    ``print` `(countDigits(n)) ` ` `

## C#

 `// C# program to find ` `// count of digits in N  ` `// factorial raised to N ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `static` `int` `countDigits(``int` `n) ` `{ ` `    ``// we take sum of logarithms  ` `    ``// as explained in the approach ` `    ``double` `ans = 0; ` `    ``for` `(``int` `i = 1; i <= n; i++) ` `        ``ans += Math.Log10(i); ` ` `  `    ``// multiply the ` `    ``// result with n ` `    ``ans = ans * n; ` `    ``return` `1 + (``int``)Math.Floor(ans); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main ()  ` `{ ` `    ``int` `n = 4; ` `    ``Console.WriteLine( ` `            ``countDigits(n) + ``"\n"``); ` `} ` `} ` ` `  `// This code is contributed ` `// by anuj_67. `

## PHP

 ` `

Output:

```6
```

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