leetcode - 172. Factorial Trailing Zeroes

Description

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 <= 10^4

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

Solution

Didn’t know how to solve until I saw the solution…

For the trailing zeroes, 2x5 makes a trailing zero. Since 2 appears more frequently than 5, the trailing zeroes depend solely on the number of 5 that appears in the factorial of n. For numbers like 25, it contains two 5s, so actually we need to find the number of multiple of 5, that is: 5, 5x5, 5x5x5, 5x5x5x5, …

Time complexity: $o(\log n)$, note the log is based on 5.
Space complexity: $o(1)$

Code

class Solution:
    def trailingZeroes(self, n: int) -> int:
        return 0 if n == 0 else n // 5 + self.trailingZeroes(n // 5)