LeetCode Solution - Problem 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:

Java:

class Solution {
    public int trailingZeroes(int n) {
       int count = 0;
       for (int i = 5; i <= n; i = i * 5) {
           count = count + n/i;
       }
       return count;
    }
}

Python:

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