182 lines
5.0 KiB
Java
182 lines
5.0 KiB
Java
import java.time.LocalDate;
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import java.time.format.DateTimeFormatter;
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import java.time.format.DateTimeParseException;
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import java.util.ArrayList;
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import java.util.List;
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import java.util.regex.Matcher;
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import java.util.regex.Pattern;
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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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if (email.contains(" "))
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return false;
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int count = 0;
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int index = 0;
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// rule 1:
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for (int i = 0; i < email.length() -1; i++) {
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if (email.charAt(i) == '@') {
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count++;
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index = i;
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}
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}
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if (count != 1)
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return false;
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// rule 2:
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String localPart = email.substring(0,index);
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String domainPart = email.substring(index + 1);
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if (localPart.isEmpty() || domainPart.isEmpty())
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return false;
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if (localPart.startsWith(".") || localPart.endsWith(".") ||
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domainPart.startsWith(".") || domainPart.endsWith("."))
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return false;
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if (localPart.contains(".."))
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return false;
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if (domainPart.contains("_"))
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return false;
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return true;
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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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return null;
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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 "!@#$%^&*"
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- has no white-space in it
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*/
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public int findValidPasswords(String string) {
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if (string.length() < 8)
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return 0;
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int validPassWords = 0;
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List<String> subStrings = List.of(string.split("\\s+"));
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for (String newPart : subStrings) {
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//at least 8 characters
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if (newPart.length() < 8)
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continue;
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// at least one uppercase letter
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if (!newPart.matches(".*[A-Z].*"))
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continue;
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// and at least a lowercase
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if (!newPart.matches(".*[a-z].*"))
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continue;
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// at least one number
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if (!newPart.matches(".*\\d.*"))
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continue;
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// at least one spacial character
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if (!newPart.matches(".*[!@#$%^&*].*"))
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continue;
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validPassWords++;
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}
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return validPassWords;
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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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List<String> list = new ArrayList<>();
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Pattern words = Pattern.compile("[a-zA-Z]+");
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Matcher matcher = words.matcher(string);
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while (matcher.find()) {
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String word = matcher.group();
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if (word.length() < 2)
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continue;
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String reversed = new StringBuilder(word).reverse().toString();
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if (reversed.equalsIgnoreCase(word))
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list.add(word);
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}
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return list;
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}
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public static void main(String[] args) {
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// you can test your code here
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BonusExercises kir = new BonusExercises();
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System.out.println(kir.findValidPasswords("""
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[09:15] Dev1: Just changed my password to CodeMaster@2025. \s
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[09:17] Dev2: Haha, mine's still qwerty123, no special chars. \s
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[09:19] Dev3: I use GitHubSuper#1 but need a better one. \s
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[09:21] Dev4: AdminPass42! is good, right? \s
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[09:23] Dev5: No, too simple. I switched to UltraSecure$99 last week. \s
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[09:25] Dev6: Wait, are we sharing passwords here? \uD83D\uDE02 \s
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"""));
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}
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}
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