# Given an integer n, return the number of trailing zeroes in n!.
# Follow up: Could you write a solution that works in logarithmic time complexity?

# 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.

class Solution:
    def trailingZeroes(self, n):
        r = 0
        while n > 0:
            n //= 5
            r += n
        return r
