929. Unique Email Addresses

Description

Every email consists of a local name and a domain name, separated by the @ sign.

For example, in [e-mail], alice is the local name, and leetcode.com is the domain name.

Besides lowercase letters, these emails may contain '.'s or '+'s.

If you add periods ('.') between some characters in the local name part of an email address, mail sent there will be forwarded to the same address without dots in the local name. For example, "[e-mail]" and "[e-mail]" forward to the same email address. (Note that this rule does not apply for domain names.)

If you add a plus ('+') in the local name, everything after the first plus sign will be ignored. This allows certain emails to be filtered, for example [e-mail] will be forwarded to [e-mail]. (Again, this rule does not apply for domain names.)

It is possible to use both of these rules at the same time.

Given a list of emails, we send one email to each address in the list. How many different addresses actually receive mails?

Constraints

• 1 <= emails[i].length <= 100

• 1 <= emails.length <= 100

• Each emails[i] contains exactly one '@' character.

• All local and domain names are non-empty.

• Local names do not start with a '+' character.

Approach

Links

• GeeksforGeeks

• Leetcode
• ProgramCreek

• YouTube

Examples

Example 1

Input: ["[e-mail]", "[e-mail]", "[e-mail]"]

Output: 2

Explanation: "[e-mail]" and "[e-mail]" actually receive mails

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int numUniqueEmails(String[] emails) {
        if(emails == null || emails.length == 0) return 0;
        
        Set<String> emailSet = new HashSet();
        for(String email: emails) {
            StringBuilder emailSB = new StringBuilder();
            int i = 0;
            char ch = email.charAt(i);
            boolean flag = true;
            while(ch != '@') {
                if(ch == '+') flag = false;
                if(flag && ch != '.') emailSB.append(ch);
                ch = email.charAt(++i);
            }
            emailSB.append(email.substring(i));
            emailSet.add(emailSB.toString());
        }
        return emailSet.size();
    }
}

Solution 2

/**
 * Time complexity : 
 * Space complexity : 
 */
 
 class Solution {
    public int numUniqueEmails(String[] emails) {
        if(emails == null || emails.length == 0) return 0;
        
        Set<String> uniqueEmails = new HashSet();
        
        for(String email: emails) {
            uniqueEmails.add(formatEmail(email));
        }
        return uniqueEmails.size();
    }
    
    private String formatEmail(String email) {
        int n = email.length();
        char[] formattedEmail = new char[n];
        int readIdx = 0, 
            writeIdx = 0;
        char ch;
        
        while((ch = email.charAt(readIdx)) != '@') {
            if(ch == '.') {
                //
            } else if(ch == '+') {
                break;
            } else {
                formattedEmail[writeIdx++] = ch;
            }
            readIdx++;
        }
        
        while((ch = email.charAt(readIdx)) != '@') {
            readIdx++;
        }
        
        while(readIdx < n) {
            formattedEmail[writeIdx++] = email.charAt(readIdx);
            readIdx++;
        }
        
        return new String(formattedEmail, 0, writeIdx);
    }
}

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