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Count digits in a factorial

Last Updated : 3 Oct, 2025

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] Calculate factorial of Number - O(n) Time and O(1) Space:

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 .

[Better approach] Using logarithmic property

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.


Output
199

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

[Best Approach] Using Stirling's approximation formula

  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 , and M_PI is the mathematical constant .
  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. 

Output
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.

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