929. Unique Email Addresses

Description

Every valid email consists of a local name and a domain name, separated by the '@' sign. Besides lowercase letters, the email may contain one or more '.' or '+'.

• For example, in "alice@leetcode.com", "alice" is the local name, and "leetcode.com" is the domain name.

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. Note that this rule does not apply to domain names.

• For example, "alice.z@leetcode.com" and "alicez@leetcode.com" forward to the same email address.

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. Note that this rule does not apply to domain names.

• For example, "m.y+name@email.com" will be forwarded to "my@email.com".

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

Given an array of strings emails where we send one email to each emails[i], return the number of different addresses that actually receive mails.

 

Example 1:

Input: emails = ["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.

Example 2:

Input: emails = ["a@leetcode.com","b@leetcode.com","c@leetcode.com"]
Output: 3

 

Constraints:

• 1 <= emails.length <= 100
• 1 <= emails[i].length <= 100
• emails[i] consist of lowercase English letters, '+', '.' and '@'.
• Each emails[i] contains exactly one '@' character.
• All local and domain names are non-empty.
• Local names do not start with a '+' character.
• Domain names end with the ".com" suffix.
• Domain names must contain at least one character before ".com" suffix.

Solutions

Solution 1: Hash Table

We can use a hash table $s$ to store all unique email addresses. Then, we traverse the array $\textit{emails}$. For each email address, we split it into the local part and the domain part. We process the local part by removing all dots and ignoring characters after a plus sign. Finally, we concatenate the processed local part with the domain part and add it to the hash table $s$.

In the end, we return the size of the hash table $s$.

The time complexity is $O(L)$, and the space complexity is $O(L)$, where $L$ is the total length of all email addresses.

Python3JavaC++GoTypeScriptRustJavaScript

class Solution:
    def numUniqueEmails(self, emails: List[str]) -> int:
        s = set()
        for email in emails:
            local, domain = email.split("@")
            t = []
            for c in local:
                if c == ".":
                    continue
                if c == "+":
                    break
                t.append(c)
            s.add("".join(t) + "@" + domain)
        return len(s)

class Solution {
    public int numUniqueEmails(String[] emails) {
        Set<String> s = new HashSet<>();
        for (String email : emails) {
            String[] parts = email.split("@");
            String local = parts[0];
            String domain = parts[1];
            StringBuilder t = new StringBuilder();
            for (char c : local.toCharArray()) {
                if (c == '.') {
                    continue;
                }
                if (c == '+') {
                    break;
                }
                t.append(c);
            }
            s.add(t.toString() + "@" + domain);
        }
        return s.size();
    }
}

class Solution {
public:
    int numUniqueEmails(vector<string>& emails) {
        unordered_set<string> s;
        for (const string& email : emails) {
            size_t atPos = email.find('@');
            string local = email.substr(0, atPos);
            string domain = email.substr(atPos + 1);
            string t;
            for (char c : local) {
                if (c == '.') {
                    continue;
                }
                if (c == '+') {
                    break;
                }
                t.push_back(c);
            }
            s.insert(t + "@" + domain);
        }
        return s.size();
    }
};

func numUniqueEmails(emails []string) int {
    s := make(map[string]struct{})
    for _, email := range emails {
        parts := strings.Split(email, "@")
        local := parts[0]
        domain := parts[1]
        var t strings.Builder
        for _, c := range local {
            if c == '.' {
                continue
            }
            if c == '+' {
                break
            }
            t.WriteByte(byte(c))
        }
        s[t.String()+"@"+domain] = struct{}{}
    }
    return len(s)
}

function numUniqueEmails(emails: string[]): number {
    const s = new Set<string>();
    for (const email of emails) {
        const [local, domain] = email.split('@');
        let t = '';
        for (const c of local) {
            if (c === '.') {
                continue;
            }
            if (c === '+') {
                break;
            }
            t += c;
        }
        s.add(t + '@' + domain);
    }
    return s.size;
}

use std::collections::HashSet;

impl Solution {
    pub fn num_unique_emails(emails: Vec<String>) -> i32 {
        let mut s = HashSet::new();

        for email in emails {
            let parts: Vec<&str> = email.split('@').collect();
            let local = parts[0];
            let domain = parts[1];
            let mut t = String::new();
            for c in local.chars() {
                if c == '.' {
                    continue;
                }
                if c == '+' {
                    break;
                }
                t.push(c);
            }
            s.insert(format!("{}@{}", t, domain));
        }

        s.len() as i32
    }
}

/**
 * @param {string[]} emails
 * @return {number}
 */
var numUniqueEmails = function (emails) {
    const s = new Set();
    for (const email of emails) {
        const [local, domain] = email.split('@');
        let t = '';
        for (const c of local) {
            if (c === '.') {
                continue;
            }
            if (c === '+') {
                break;
            }
            t += c;
        }
        s.add(t + '@' + domain);
    }
    return s.size;
};

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