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In this approach, we use regular expressions to check for valid emails. The regular expression will ensure that the email format adheres to the specified rules: the prefix starts with a letter and consists only of permitted characters, followed by '@leetcode.com'. This method efficiently checks the constraints by leveraging regular expressions, which are well-suited for pattern matching tasks.
The time complexity of this solution is O(n), where n is the number of users, as each email is scanned once. The space complexity is O(n) as we store all valid users in a list.
1import
This Java solution uses the regex library to compile a pattern matching valid email addresses and filters valid emails by iterating over the user list. The 'matches' method is used to verify if each email satisfies the pattern.
This approach involves manually parsing the email to check the prefix and domain components. We split the email at the '@' character and validate both parts separately. This is an alternative to regex that provides a more step-by-step checking method.
The time complexity is O(n * m), where n is the number of users and m is the length of the largest email string due to manual character checking. The space complexity is O(n) for the result list.
1import java.util.*;
2
3public class ValidEmailsCheck {
4 public static List<Map<String, String>> findValidEmails(List<Map<String, String>> users) {
5 List<Map<String, String>> validEmails = new ArrayList<>();
6
7 for (Map<String, String> user : users) {
8 String email = user.get("mail");
9 if (isValidEmail(email)) {
10 validEmails.add(user);
11 }
12 }
13 return validEmails;
14 }
15
16 private static boolean isValidEmail(String email) {
17 if (!email.contains("@leetcode.com")) {
18 return false;
19 }
20 String[] parts = email.split("@", 2);
21 if (parts.length != 2 || !parts[1].equals("leetcode.com")) {
22 return false;
23 }
24
25 String prefix = parts[0];
26 if (prefix.isEmpty() || !Character.isLetter(prefix.charAt(0))) {
27 return false;
28 }
29 for (char c : prefix.toCharArray()) {
30 if (!Character.isLetterOrDigit(c) && c != '.' && c != '_' && c != '-') {
31 return false;
32 }
33 }
34 return true;
35 }
36
37 public static void main(String[] args) {
38 List<Map<String, String>> users = new ArrayList<>();
39 users.add(new HashMap<>(Map.of("user_id", "1", "name", "Winston", "mail", "winston@leetcode.com")));
40 users.add(new HashMap<>(Map.of("user_id", "2", "name", "Jonathan", "mail", "jonathanisgreat")));
41 users.add(new HashMap<>(Map.of("user_id", "3", "name", "Annabelle", "mail", "bella-@leetcode.com")));
42 users.add(new HashMap<>(Map.of("user_id", "4", "name", "Sally", "mail", "sally.come@leetcode.com")));
43 users.add(new HashMap<>(Map.of("user_id", "5", "name", "Marwan", "mail", "quarz#2020@leetcode.com")));
44 users.add(new HashMap<>(Map.of("user_id", "6", "name", "David", "mail", "david69@gmail.com")));
45 users.add(new HashMap<>(Map.of("user_id", "7", "name", "Shapiro", "mail", ".shapo@leetcode.com")));
46
47 List<Map<String, String>> validUsers = findValidEmails(users);
48 System.out.println(validUsers);
49 }
50}This Java solution manually parses the email to separate the prefix and domain parts. It checks if the domain matches '@leetcode.com' and validates the prefix according to rules. This approach foregoes regex for explicit parsing and logical checking.