Description
https://leetcode.com/problems/factorial-trailing-zeroes/
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.
Explanation
one 5 one zero
Python Solution
class Solution:
def trailingZeroes(self, n: int) -> int:
product = 1
zero_count = 0
while n > 1:
zero_count += n // 5;
n //= 5
return zero_count- Time complexity: ~N
- Space complexity: ~1
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