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Factorial Trailing Zeroes
Given an integer n, return the number of trailing zeroes in n!.
Given an integer n, return the number of trailing zeroes in n!.
Note that n! = n * (n - 1) * (n - 2) * ... * 3 * 2 * 1.
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.
Example 3
Input: n = 0
Output: 0
Constraints
- 0 <= n <= 10^4
Follow-up
Could you write a solution that works in logarithmic time complexity?
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