[Solved] Given an integer n, return the number of trailing zeroes in n!. Note that n! = n * (n – 1) * (n – 2) * … * 3 * 2 * 1.

Question

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?

Python Solution

class Solution:
    def trailingZeroes(self, n: int) -> int:
        zc=0
        while n>0:
            n//=5
            zc+=n
        return zc