Given an integer n, return the number of trailing zeroes in n!.
Follow up: Could you write a solution that works in logarithmic time complexity?
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
function trailingZeroes(n: number): number {
let count = 0;
while (n > 0) {
n = Math.floor(n / 5);
count += n;
}
return count;
}