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?
class Solution:
     def trailingZeroes(self, n):
         """
         :type n: int
         :rtype: int
         """
         c=0
         while n>0:
             n//=5
             c+=n
         return c