Unique Email Addresses LeetCode Solution  

Question

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 email[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

Explanation

This question is easy but looks a bit complicated at first sight. The problem can be solved using hashmap, set, unordered set. Lets see the approach now. Dividing the each email into two parts

Local name - Building it by traversing the each character in every email and adding it into a tempMail string

i) if local name contains "." continue making the local name in the tempMail string.

ii) if local name contains "+" everything after would be ignored thus break the loop.

iii) if while traversing "@" is encountered that means we have completed the traversal for local name, break the loop.

Domain Name - No filters are given for this, So just adding it as it after we have made our local string.

An unordered_set of strings could be used to store the every new tempMail. So that even though we will be pushing the multiple copy entries, its size would only give the size of unique entries. In this piece of code, I took lots of times to recall regex, and it is the most unreadable part, so let me explain:

/\+.*$|\./g

+ is a special char, so adds \.

.* means any character after +.

$, in regex, it represents the end of string.

| equals to or.

. is also a special char, so adds \.

In the end of regex, g, global search, means finding all matches in input and replace them.

In sum, replace the substring that after sign + or the char . with empty string.

That's all you need to know about. Hope you understood the explanation. I recommend to try the problem by yourself then if you are stuck you can visit the solution given below in Java, C++, Python.

Program Code

Java

class Solution {
    public int numUniqueEmails(String[] emails) {
        HashSet<String> set = new HashSet<>();
        for(String s : emails){
            StringBuilder sb = new StringBuilder();
            int at = s.indexOf("@");
            for(int i=0;i<at;i++){
                char ch = s.charAt(i);
                if(ch == '+') break;
                if(ch!='.') sb.append(ch);
            }
            sb.append(s.substring(at));
            set.add(sb.toString());
        }
        return set.size();
    }
}

C++

int numUniqueEmails(vector<string>& emails) {
    set<string>s;
    for(auto i:emails){
        string d=i;
        string temp;
        int x=0;
        for(x;d[x]!='@';x++){
            if(d[x]=='+'){
                break;
                
            }
            
            if(d[x]!='.'){
                temp.push_back(d[x]);
                
            }
        }
        while(d[x]!='@'){
            x++;
        }
        for(x;x<d.size();x++){
            temp.push_back(d[x]);
        }
     
        cout<<temp<<endl;
        s.insert(temp);
    }
    return s.size();
    
}

Python

class Solution:
    def numUniqueEmails(self, emails: List[str]) -> int:
        email_set = set()
        for email in emails:
            name, domain = email.split('@')
            name = name.replace('.', '')
            
            if '+' in name:
                name = name[:name.index('+')]
                
            email_set.add(name + '@' + domain)
        
        return len(email_set)

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