Problem Statement
Given an integer n, return the number of trailing zeroes in n!.
Example 1:
Input: 3 Output: 0 Explanation: 3! = 6, no trailing zero.
Example 2:
Input: 5 Output: 1 Explanation: 5! = 120, one trailing zero.
Note: Your solution should be in logarithmic time complexity.
Solution:
class Solution:
def trailingZeroes(self, n: int) -> int:
if not n:
return 0
counter = 0
while n>=5:
counter += n//5
n //=5
return counter
