# Solving Leetcode Interviews in Seconds with AI: Factorial Trailing Zeroes


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "172" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an integer n, return the number of trailing zeroes in n!. Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1.   Example 1:  Input: n = 3 Output: 0 Explanation: 3! = 6, no trailing zero.  Example 2:  Input: n = 5 Output: 1 Explanation: 5! = 120, one trailing zero.  Example 3:  Input: n = 0 Output: 0    Constraints:  0 <= n <= 104    Follow up: Could you write a solution that works in logarithmic time complexity? 

	# Explanation
	Here's the approach to solve this efficiently:

*   Trailing zeros in a factorial are determined by the number of times 5 appears as a factor. This is because the number of 2s will always be greater than or equal to the number of 5s, and a trailing zero is formed by a pair of 2 and 5.
*   We can count the number of factors of 5 by repeatedly dividing `n` by 5 and summing the quotients. This accounts for multiples of 5, multiples of 25 (which contribute an extra factor of 5), multiples of 125, and so on.
*   The process stops when `n` divided by 5 becomes zero.

*   Runtime Complexity: O(log<sub>5</sub>(n)). Storage Complexity: O(1).

	
	# Code
	```python
	def trailingZeroes(n: int) -> int:
    """
    Given an integer n, return the number of trailing zeroes in n!.
    """
    count = 0
    while n > 0:
        n //= 5
        count += n
    return count
	```