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# Factorial Trailing Zeroes LeetCode Solution

## 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:

• `0 <= n <= 104`

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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