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.

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