import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class BonusExercises {

    /*
        complete the method below, so it will validate an email address
        1. Must have exactly one @ (no more, no less)
        2. Split into local-part and domain (before @ and after @)
        3. Local-part rules:
            - Can't be empty
            - Can't start or end with dot
            - Can't have two dots in a row
        4. Domain rules:
            - Can't be empty
            - Can't start or end with hyphen
            - Can't have underscores
            - Each segment (between dots) must follow same hyphen rules
     */
    public boolean validateEmail(String email) {
        String regex = "^[a-zA-z0-9_+-]+(?:\\.[a-zA-Z0-9_+-]+)*@(?:[a-zA-Z0-9-]+\\.)+[a-zA-Z0-9-]+$";
        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(email);

        return matcher.matches();
    }

    /*
        This method should find and return the first date in a string.

        Supported formats:
        - American:   MM/DD/YYYY   (e.g., 12/09/2023)
        - British:    DD/MM/YYYY   (e.g., 12/09/2023 — same pattern, context matters)
        - ISO:        YYYY-MM-DD   (e.g., 2024-07-15)
        - Slash variant: YYYY/MM/DD (e.g., 2025/01/01)

        If no match for a date is found in the string, return null.
    */
    public String findDate(String string) {
        String regex1 = "(0[1-9]|1[0-2])\\/(0[1-9]|[12][0-9]|30)\\/\\d{4}|(0[1-9]|[12][0-9]|30)\\/(0[1-9]|1[0-2])\\/\\d{4}";
        String regex2 = "(\\d{4}-(0[1-9]|1[0-2])-(0[1-9]|[12][0-9]|30))";
        String regex3 = "(\\d{4}\\/(0[1-9]|1[0-2])\\/(0[1-9]|[12][0-9]|30))";
        Pattern pattern = Pattern.compile(regex1 + "|" + regex2 + "|" + regex3);
        Matcher matcher = pattern.matcher(string);

        if(matcher.find()){
            return matcher.group();
        }
        return null;
    }

    /*
        given a string, implement the method to detect all valid passwords
        then, it should return the count of them

        a valid password has the following properties:
        - at least 8 characters
        - has to include at least one uppercase letter, and at least a lowercase
        - at least one number and at least a special char "!@#$%^&*"
        - has no white-space in it
     */
    public int findValidPasswords(String string) {
        String regex = "^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[!@#$%^&*]).{8,}$";
        Pattern pattern = Pattern.compile(regex);

        String[] parts = string.split("\\s+");
        int counter = 0;
        for(String part : parts){
            Matcher matcher = pattern.matcher(part);
            if(matcher.matches()){
                counter++;
            }
        }
        return counter;
    }

    /*
        you should return a list of *words* which are palindromic
        by word we mean at least 3 letters with no whitespace in it

        note: your implementation should be case-insensitive, e.g. Aba -> is palindrome
     */
    public List<String> findPalindromes(String string) {
        List<String> list = new ArrayList<>();
        // todo
        return list;
    }

    public static void main(String[] args) {
        // you can test your code here
    }
}