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?

SOLUTION:

class Solution:
    def trailingZeroes(self, n: int) -> int:
        ctr = [0, 0]
        for i in range(1, n + 1):
            curr = i
            while curr % 2 == 0:
                ctr[0] += 1
                curr = curr // 2
            while curr % 5 == 0:
                ctr[1] += 1
                curr = curr // 5
        return min(ctr)