Count digits in a factorial using Logarithm

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:  5
Output: 3
Explanation: 5! = 120, i.e., 3 digits

Input: 10
Output: 7
Explanation: 10! = 3628800, i.e., 7 digits
 

Naive approach: To solve the problem follow the below idea:

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!). 

Count digits in a factorial using the property of logarithms:

To solve the problem follow the below idea:

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 the implementation of the above approach:

// A C++ program to find the number of digits in // a factorial #include <bits/stdc++.h> 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() {  // Function call  cout << findDigits(1) << endl;  cout << findDigits(5) << endl;  cout << findDigits(10) << endl;  cout << findDigits(120) << endl;  return 0; } 

// 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)  {  // Function call  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 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 if __name__ == "__main__": print(findDigits(1)) print(findDigits(5)) print(findDigits(10)) print(findDigits(120)) # This code is contributed by mits 

// 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()  {  // Function call  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 

// A Javascript program to find the number of digits in  // a factorial  // This function receives an integer n, and returns  // the number of digits present in n!  function 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   let digits = 0;   for (let i=2; i<=n; i++)   digits += Math.log10(i);   return Math.floor(digits) + 1;  }  // Driver code     document.write(findDigits(1) + "<br>");   document.write(findDigits(5) + "<br>");   document.write(findDigits(10) + "<br>");   document.write(findDigits(120) + "<br>");    //This code is contributed by Mayank Tyagi 

<?php // PHP program to find  // the number of digits  // in a factorial // This function receives  // an integer n, and returns // the number of digits present in n! function 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 = 2; $i <= $n; $i++) $digits += log10($i); return floor($digits) + 1; } // Driver code // Function call echo findDigits(1), "\n"; echo findDigits(5), "\n"; echo findDigits(10), "\n"; echo findDigits(120), "\n"; // This code is contributed by Ajit. ?> 

1 3 7 199

Time complexity: O(N log N) since calculating log in a loop
Auxiliary space: O(1) because it is using constant variables

 Approach 2: Using Stirling's approximation formula to calculate the factorial and logarithm to count the number of digits.

1. The countDigitsInFactorial(int n) function takes an integer n as input and returns the number of digits in the factorial of n. If n is negative, it returns 0. If n is 0 or 1, the factorial is 1, and it returns 1.
2. In the countDigitsInFactorial(int n) function, the double x variable is declared and initialized using the Stirling's approximation formula for the factorial. This formula provides a good approximation of the value of the factorial for large values of n. 
3. where M_E is the mathematical constant Euler number e, and M_PI is the mathematical constant \pi.
4. The formula used in this code is a simplified version of Stirling's approximation that takes the logarithm of the above formula to get the number of digits in the factorial. 

#include <cmath> #include <iostream> using namespace std; int countDigitsInFactorial(int n) {  if (n < 0) {  return 0;  }  if (n <= 1) {  return 1;  }  double x  = (n * log10(n / M_E) + log10(2 * M_PI * n) / 2.0);  return floor(x) + 1; } int main() {  cout << countDigitsInFactorial(1) << endl;  cout << countDigitsInFactorial(5) << endl;  cout << countDigitsInFactorial(10) << endl;  cout << countDigitsInFactorial(120) << endl;  return 0; } 

// Java implementation of above approach import java.lang.Math; import java.util.Scanner; public class Solution { public static int countDigitsInFactorial(int n) {    // factorial exists only for n>=0 if (n < 0) { return 0; } if (n <= 1) { return 1; }  // Calculating the digit's values double x = (n * Math.log10(n / Math.E) + Math.log10(2 * Math.PI * n) / 2.0);  // returning the floor value + 1 return (int) Math.floor(x) + 1;  } public static void main(String[] args) {    // calling the countDigitInFactorial function  System.out.println(countDigitsInFactorial(1));  System.out.println(countDigitsInFactorial(5));  System.out.println(countDigitsInFactorial(10));  System.out.println(countDigitsInFactorial(120)); } } 

import math def countDigitsInFactorial(n): if n < 0: return 0 if n <= 1: return 1 # Using Stirling's approximation formula to count the number of digits x = (n * math.log10(n / math.e) + math.log10(2 * math.pi * n) / 2.0) # Floor the result of the formula and add 1 to get the number of digits return math.floor(x) + 1 # Testing the function with sample inputs print(countDigitsInFactorial(1)) print(countDigitsInFactorial(5)) print(countDigitsInFactorial(10)) print(countDigitsInFactorial(120)) 

using System; public class Program {  static int CountDigitsInFactorial(int n)  {  if (n < 0) {  return 0;  }  if (n <= 1) {  return 1;  }  double x = (n * Math.Log10(n / Math.E)  + Math.Log10(2 * Math.PI * n) / 2.0);  return (int)Math.Floor(x) + 1;  }  public static void Main()  {  Console.WriteLine(CountDigitsInFactorial(1));  Console.WriteLine(CountDigitsInFactorial(5));  Console.WriteLine(CountDigitsInFactorial(10));  Console.WriteLine(CountDigitsInFactorial(120));  } } 

// Javascript program for the above approach // Function to count digits in factorials // of the number n function countDigitsInFactorial(n) {  if (n < 0) {  return 0;  }  if (n <= 1) {  return 1;  }  let x = n * Math.log10(n / Math.E) + Math.log10(2 * Math.PI * n) / 2.0;  return Math.floor(x) + 1; } // Driver Code console.log(countDigitsInFactorial(1)); console.log(countDigitsInFactorial(5)); console.log(countDigitsInFactorial(10)); console.log(countDigitsInFactorial(120)); 

1 3 7 199

Time complexity: O(1)

The time complexity of the above approach to count the number of digits in n! using Stirling's approximation and logarithms is O(1), meaning it is constant time complexity.
Auxiliary space: O(1)

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