content: package LeetCode.Easy.FactorialTrailingZeroes;
  
  /*
    172. Factorial Trailing Zeroes
  
    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
  */
  
  public class Solution {
    public int trailingZeroes(int n) {
      if (n == 0) return 0;
  
      int count = 0;
  
      for (int i = 5; n / i >= 1; i *= 5) count += Math.floor(n / i);
  
      return count;
    }
  
    public static void main(String[] args) {
      Solution solution = new Solution();
  
      // should be 0
      System.out.println(solution.trailingZeroes(3));
  
      // should be 1
      System.out.println(solution.trailingZeroes(5));
  
      // should be 0
      System.out.println(solution.trailingZeroes(0));
  
      // should be 7
      System.out.println(solution.trailingZeroes(30));
    }
  }
  