<p>Given an integer <code>n</code>, return <em>the number of trailing zeroes in </em><code>n!</code>.</p>

<p>Note that <code>n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> 3! = 6, no trailing zero.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 5
<strong>Output:</strong> 1
<strong>Explanation:</strong> 5! = 120, one trailing zero.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> n = 0
<strong>Output:</strong> 0
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= n &lt;= 10<sup>4</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you write a solution that works in logarithmic time complexity?</p>

class Solution:
    def trailingZeroes(self, n: int) -> int:
        count = 0
        for i in range(n, 0, -1):
            while i % 5 == 0:
                i //= 5
                count += 1
        return count
16400