Medium
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?
function trailingZeroes(n: number): number {
let base = 5
let count = 0
while (n >= base) {
count += Math.floor(n / base)
base *= 5
}
return count
}
export { trailingZeroes }