GeeksforGeeks: Count the Number of Digits in a Factorial | Efficient Solution Explained

In this video, we’ll tackle the GeeksforGeeks problem of counting the number of digits in the factorial of a given number
𝑁
N. Instead of computing the factorial directly, which can be enormous, we'll use an efficient logarithmic approach to determine the number of digits. This method is particularly useful for handling large numbers and is commonly used in coding interviews.

Problem Statement:
Given an integer 𝑁
N, you need to find the number of digits that appear in its factorial. The factorial of 𝑁
N is defined as 𝑁!
N! where:
𝑁!=1×2×3×…×𝑁

Examples:

Example 1:
Input:
𝑁 = 5
N = 5
Output: 3
Explanation:
5 != 120, which has 3 digits.

Example 2:
Input:
𝑁 = 10
Output: 7
Explanation:
10 != 3628800
10!=3628800, which has 7 digits.
Your Task:
You need to implement the function facDigits() that takes an integer 𝑁
N as input and returns the number of digits in the factorial of 𝑁.

### Solution Approach

**Efficient Approach:**
Given that the factorial of even moderately large numbers can result in extremely large values, directly computing the factorial and then counting the digits is impractical. Instead, we use properties of logarithms to determine the number of digits in the factorial of a number.

**Mathematical Insight:**
The number of digits \(d\) in a number \(x\) can be found using the formula:
\[d = \lfloor \log_{10}(x) \rfloor + 1\]

For the factorial of a number \(N\), \(N!\), we can use the properties of logarithms:
\[\log_{10}(N!) = \log_{10}(1) + \log_{10}(2) + \ldots + \log_{10}(N)\]

So, the number of digits in \(N!\) is given by:
\[d = \lfloor \log_{10}(N!) \rfloor + 1\]

By summing up the logarithms of all integers from 1 to \(N\), we can efficiently calculate the number of digits in \(N!\).

GitHub Solution Repository Link:

Thanks for tuning in! I hope this video helps you master the technique for counting digits in a factorial efficiently. If you enjoyed the content, please like, share, and subscribe for more programming challenges and solutions. Keep coding and see you in the next video!

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