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, (N!)^N

Examples:

Input: 4
Output: 6
Explanations: (4!)^4 = (24)^4 = 331776. 
              Total number of digits in 331776 is 6.

Input: 5
Output: 11
Explanations: (5!)^5 = (120)^5 = 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.

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 (N!)^N 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 N * log10(N!).
 
and we know, N! = 1 * 2 * 3 * 4 *....* N.
Therefore we can reduce further.
 
N * log10(N!)
 
= N * log10(1 * 2 * 3 * 4 *....* N)
 
= N * [log10 (1) + log10 (2) + log10 (3) + .... + log10 (N)]
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++

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// CPP program to find count of digits in N 
// factorial raised to N
#include <bits/stdc++.h>
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;
}

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Java

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// 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.

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Python3

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# 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))
  

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C#

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// 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.

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PHP

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<?php
// PHP program to find count of 
// digits in N factorial raised to N 
function countDigits ($n
    // we take sum of logarithms as
    // explained in the approach 
    $ans = 0; 
    for ($i = 1; $i <= $n; $i++) 
        $ans += log10($i); 
  
    // multiply the result with n 
    $ans = $ans * $n
    return 1 + floor($ans); 
  
// Driver Code
$n = 4; 
echo countDigits($n) , "\n"
  
// This code is contributed 
// by jit_t    
?>

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Output:

6


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Improved By : vt_m, jit_t, Ita_c