human
I found an interesting problem on https://leetcode.com/problems/factorial-trailing-zeroes/:
Given an integer n, return the number of trailing zeroes in n!.

Example 1:

Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.

Example 2:

Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.

Note: Your solution should be in logarithmic time complexity.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
     def trailingZeroes(self, n):
         res = 5
         ans = 0
         while res < n+1:
             ans += int(n/res)
             res = 5*res
         return ans