Geek For Geeks | Trailing Zeroes in a factorial of a number | Java

For an integer N find the number of trailing zeroes in N!.

Example 1:

Input:
N = 5
Output:
1
Explanation:
5! = 120 so the number of trailing zero is 1.

Expected Time Complexity: O(logN)
Expected Auxiliary Space: O(1)
Constraints:
1 <= N <= 109

  public class TrailingZeroes {

    public static void main(String[] args) {
       int n = 384;
       int count = 0;

        for (int i = 5; n/i >= 1 ; i *= 5) {
            count += n/i;
        }


        System.out.println(count);
    }
}