## Problem – Factorial Trailing Zeroes

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

**Example 2:**

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

**Example 3:**

```
Input: n = 0
Output: 0
```

**Constraints:**

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

### Factorial Trailing Zeroes LeetCode Solution in Java

```
return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
```

### Factorial Trailing Zeroes LeetCode Solution in C++

```
return n == 0 ? 0 : n / 5 + trailingZeroes(n / 5);
```

### Factorial Trailing Zeroes LeetCode Solution in Python

```
return 0 if n == 0 else n / 5 + self.trailingZeroes(n / 5)
```

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