implement Bonus exercises
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@@ -5,68 +5,84 @@ import java.util.regex.Pattern;
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public class BonusExercises {
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public class BonusExercises {
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/*
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complete the method below, so it will validate an email address
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1. Must have exactly one @ (no more, no less)
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2. Split into local-part and domain (before @ and after @)
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3. Local-part rules:
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- Can't be empty
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- Can't start or end with dot
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- Can't have two dots in a row
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4. Domain rules:
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- Can't be empty
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- Can't start or end with hyphen
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- Can't have underscores
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- Each segment (between dots) must follow same hyphen rules
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*/
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public boolean validateEmail(String email) {
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public boolean validateEmail(String email) {
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String regex = ""; // todo
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String regex = "^(?!\\.)(?!.*\\.\\..*)([0-9A-Za-z._]+)(?<!\\.)@(?!-)([0-9A-Za-z.\\-]+)(?<!-)$";
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Pattern pattern = Pattern.compile(regex);
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Pattern pattern = Pattern.compile(regex);
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Matcher matcher = pattern.matcher(email);
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Matcher matcher = pattern.matcher(email);
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return matcher.matches();
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return matcher.matches();
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}
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}
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/*
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This method should find and return the first date in a string.
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Supported formats:
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- American: MM/DD/YYYY (e.g., 12/09/2023)
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- British: DD/MM/YYYY (e.g., 12/09/2023 — same pattern, context matters)
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- ISO: YYYY-MM-DD (e.g., 2024-07-15)
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- Slash variant: YYYY/MM/DD (e.g., 2025/01/01)
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If no match for a date is found in the string, return null.
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*/
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public String findDate(String string) {
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public String findDate(String string) {
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// todo
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String regex = "(\\b((1[0-2])|(0[1-9]))\\b/\\b((0[1-9])|([12][0-9])|(3[01]))\\b/\\b(\\d{4})\\b)|(\\b((0[1-9])|([12][0-9])|(3[01]))\\b/\\b((1[0-2])|(0[1-9]))\\b/\\b(\\d{4})\\b)|(\\b(\\d{4})\\b-\\b((1[0-2])|(0[1-9]))\\b-\\b((0[1-9])|([12][0-9])|(3[01]))\\b)|(\\b(\\d{4})\\b/\\b((1[0-2])|(0[1-9]))\\b/\\b((0[1-9])|([12][0-9])|(3[01]))\\b)";
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Pattern pattern = Pattern.compile(regex);
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Matcher matcher = pattern.matcher(string);
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if (matcher.find()) {
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return matcher.group();
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}
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return null;
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return null;
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}
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}
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/*
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given a string, implement the method to detect all valid passwords
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then, it should return the count of them
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a valid password has the following properties:
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- at least 8 characters
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- has to include at least one uppercase letter, and at least a lowercase
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- at least one number and at least a special char "!@#$%^&*"
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- has no white-space in it
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*/
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public int findValidPasswords(String string) {
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public int findValidPasswords(String string) {
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// todo
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String passwordRegex = "(?=.*[A-Z])(?=.*[a-z])(?=.*\\d)(?=.*\\W).{8,}";
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return -1;
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String tokenRegex = "(?<!\\S)[A-Za-z0-9!@#$%^&*_]{8,}(?!\\S)";
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Pattern tokenPattern = Pattern.compile(tokenRegex);
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Pattern passwordPattern = Pattern.compile(passwordRegex);
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Matcher matcher = tokenPattern.matcher(string);
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int count = 0;
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while (matcher.find()) {
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String token = matcher.group();
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Matcher passwordMatcher = passwordPattern.matcher(token);
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if (passwordMatcher.matches()){
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count++;
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}
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}
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return count;
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}
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}
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/*
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you should return a list of *words* which are palindromic
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by word we mean at least 3 letters with no whitespace in it
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note: your implementation should be case-insensitive, e.g. Aba -> is palindrome
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*/
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public List<String> findPalindromes(String string) {
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public List<String> findPalindromes(String string) {
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List<String> list = new ArrayList<>();
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List<String> list = new ArrayList<>();
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// todo
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String[] inputList = string.split("\\W+");
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for (String word : inputList)
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{
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String regex = "";
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int wordLength = word.length();
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if (wordLength < 3) continue;
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for (int i = 0; i < wordLength/2; i++) {
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regex += "(.)";
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}
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if (wordLength % 2 != 0)
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{
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regex += ".";
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}
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for (int i = wordLength/2; i > 0; i--) {
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regex += "\\" + i;
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}
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Pattern pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
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Matcher matcher = pattern.matcher(word);
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if (matcher.matches())
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{
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list.add(word);
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}
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}
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return list;
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return list;
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}
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}
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