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172. Factorial Trailing Zeroes

Difficulty Topics

Description

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

factorial-trailing-zeroes.java
public class Solution {
    public int trailingZeroes(int n) {
        int count = 0;
        while (n != 0) {
            int tmp = n / 5;
            count += tmp;
            n = tmp;
        }
        return count;
    }
}