## Question

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

**Example 2:**

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

**Example 3:**

Input:n = 0Output:0

**Constraints:**

`0 <= n <= 10`

^{4}

**Follow up:** Could you write a solution that works in logarithmic time complexity?

## Python Solution

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