implement Bonus exercises

This commit is contained in:
Ramtin Jafari
2026-07-06 04:52:32 +03:30
parent dd399fd00a
commit fa7227822b
+60 -44
View File
@@ -5,68 +5,84 @@ 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 = ""; // todo
String regex = "^(?!\\.)(?!.*\\.\\..*)([0-9A-Za-z._]+)(?<!\\.)@(?!-)([0-9A-Za-z.\\-]+)(?<!-)$";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(email);
return matcher.matches();
}
/*
This method should find and return the first date in a string.
Supported formats:
- American: MM/DD/YYYY (e.g., 12/09/2023)
- British: DD/MM/YYYY (e.g., 12/09/2023 — same pattern, context matters)
- ISO: YYYY-MM-DD (e.g., 2024-07-15)
- Slash variant: YYYY/MM/DD (e.g., 2025/01/01)
If no match for a date is found in the string, return null.
*/
public String findDate(String string) {
// todo
String regex = "(\\b((1[0-2])|(0[1-9]))\\b/\\b((0[1-9])|([12][0-9])|(3[01]))\\b/\\b(\\d{4})\\b)|(\\b((0[1-9])|([12][0-9])|(3[01]))\\b/\\b((1[0-2])|(0[1-9]))\\b/\\b(\\d{4})\\b)|(\\b(\\d{4})\\b-\\b((1[0-2])|(0[1-9]))\\b-\\b((0[1-9])|([12][0-9])|(3[01]))\\b)|(\\b(\\d{4})\\b/\\b((1[0-2])|(0[1-9]))\\b/\\b((0[1-9])|([12][0-9])|(3[01]))\\b)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);
if (matcher.find()) {
return matcher.group();
}
return null;
}
/*
given a string, implement the method to detect all valid passwords
then, it should return the count of them
a valid password has the following properties:
- at least 8 characters
- has to include at least one uppercase letter, and at least a lowercase
- at least one number and at least a special char "!@#$%^&*"
- has no white-space in it
*/
public int findValidPasswords(String string) {
// todo
return -1;
String passwordRegex = "(?=.*[A-Z])(?=.*[a-z])(?=.*\\d)(?=.*\\W).{8,}";
String tokenRegex = "(?<!\\S)[A-Za-z0-9!@#$%^&*_]{8,}(?!\\S)";
Pattern tokenPattern = Pattern.compile(tokenRegex);
Pattern passwordPattern = Pattern.compile(passwordRegex);
Matcher matcher = tokenPattern.matcher(string);
int count = 0;
while (matcher.find()) {
String token = matcher.group();
Matcher passwordMatcher = passwordPattern.matcher(token);
if (passwordMatcher.matches()){
count++;
}
}
return count;
}
/*
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
String[] inputList = string.split("\\W+");
for (String word : inputList)
{
String regex = "";
int wordLength = word.length();
if (wordLength < 3) continue;
for (int i = 0; i < wordLength/2; i++) {
regex += "(.)";
}
if (wordLength % 2 != 0)
{
regex += ".";
}
for (int i = wordLength/2; i > 0; i--) {
regex += "\\" + i;
}
Pattern pattern = Pattern.compile(regex, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(word);
if (matcher.matches())
{
list.add(word);
}
}
return list;
}