Unique Email Address

Easy

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

For example, inalice@leetcode.com, aliceis the local name, andleetcode.comis the domain name.

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

If you add periods ('.') between some characters in thelocal namepart of an email address, mail sent there will be forwarded to the same address without dots in the local name. For example,"alice.z@leetcode.com"and"alicez@leetcode.com"forward to the same email address. (Note that this rule does not apply for domain names.)

If you add a plus ('+') in thelocal name, everything after the first plus sign will be ignored. This allows certain emails to be filtered, for example m.y+name@email.com will be forwarded to my@email.com. (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 ofemails, we send one email to each address in the list. How many different addresses actually receive mails?

Example 1:

Input: 
["test.email+alex@leetcode.com","test.e.mail+bob.cathy@leetcode.com","testemail+david@lee.tcode.com"]
Output: 
2
Explanation:
 "
testemail@leetcode.com" and "testemail@lee.tcode.com" 
actually receive mails

Note:

  • 1 <= emails[i].length <= 100
  • 1 <= emails.length <= 100
  • Each emails[i]contains exactly one'@'character.

Analysis

基本思想就是将email都转化为统一的形式canonical form,即去掉'.', '+等字符的影响,最后返回unique元素hashset的大小即可。

Solution

LeetCode official - Using indexOf(), replaceAll() - (39 ms, faster than 34.09%)

class Solution {
    public int numUniqueEmails(String[] emails) {
        Set<String> seen = new HashSet();
        for (String email: emails) {
            int i = email.indexOf('@');
            String local = email.substring(0, i);
            String rest = email.substring(i);
            if (local.contains("+")) {
                local = local.substring(0, local.indexOf('+'));
            }
            local = local.replaceAll(".", "");
            seen.add(local + rest);
        }

        return seen.size();
    }
}

Character by character, StringBuilder - (25 ms, faster than 72.51%)

class Solution {
    public int numUniqueEmails(String[] emails) {
        if (emails == null) {
            return 0;
        }
        HashSet<String> set = new HashSet<>();
        for (String email : emails) {
            set.add(convertEmailAddress(email));
        }
        return set.size();
    }

    private String convertEmailAddress(String email) {
        StringBuilder sb = new StringBuilder();

        int i = 0;
        boolean beforeAtSign = true;
        while (i < email.length()) {
            if (email.charAt(i) == '@') {
                i++;
                beforeAtSign = false;
            } else if (beforeAtSign && email.charAt(i) == '.') {
                i++;
            } else if (beforeAtSign && email.charAt(i) == '+') {
                while (i < email.length()) {
                    i++;
                    if (email.charAt(i) == '@') {
                        beforeAtSign = false;
                        break;
                    }
                }
            }
            sb.append(email.charAt(i));
            i++;
        }
        return sb.toString();
    }
}

(Preferred*) Another Character by character implementation by @FLAGbigoffer- (17ms, 89.80%)

We can do BETTER without using replace(). Actually in some situations, we even don't need scan the entire email address. Consider this case:

a+nadgkj.ansjfnakjn.gaskjgb.ubeijbg.kjncna.oskenijb.golkn.asignoaeroib.gjoibv.oanod.safa.e.f.asdf.sa.df.a@g.com

. What we do here is to scan the email address until meeting the first '+' sign. Then scan reversely to find the '@' sign.

class Solution {
    public int numUniqueEmails(String[] emails) {
        Set<String> emailsSet = new HashSet<>();

        for (String e : emails) {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < e.length(); i++) {
                if (e.charAt(i) == '.') {
                    continue;
                } else if (e.charAt(i) == '+') {
                    int idx = e.length() - 1;
                    while (e.charAt(idx) != '@') {
                        idx--;
                    }
                    sb.append(e.substring(idx));
                    break;
                } else {
                    sb.append(e.charAt(i));
                }
            }
            emailsSet.add(sb.toString());
        }

        return emailsSet.size();
    }
}

Reference

https://leetcode.com/problems/unique-email-addresses/solution/

https://leetcode.com/problems/unique-email-addresses/discuss/186798/Java-7-liner-with-comment.

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