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, 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(log₅(n)). Storage Complexity: O(1).

Code

    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