Unique Email Addresses

Unique Email Addresses - Problem

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.

Input & Output

Example 1 — Multiple Normalization Rules

$ Input: emails = ["test.email+tag@leetcode.com","test.e.mail@leetcode.com","test.email.leet+tag@leetcode.com"]

› Output: 2

💡 Note: After normalization: "test.email+tag@leetcode.com" becomes "testemail@leetcode.com", "test.e.mail@leetcode.com" becomes "testemail@leetcode.com" (same as first), and "test.email.leet+tag@leetcode.com" becomes "testemaillet@leetcode.com". So 2 unique addresses.

Example 2 — Only Dot Normalization

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

› Output: 3

💡 Note: No dots or plus signs to normalize, so all three emails remain unique: "a@leetcode.com", "b@leetcode.com", "c@leetcode.com".

Example 3 — Plus Sign Filtering

$ Input: emails = ["alice+work@example.com","alice+personal@example.com","alice@example.com"]

› Output: 1

💡 Note: All emails normalize to "alice@example.com" because everything after the plus sign in the local part is ignored.

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

Visualization

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Asked in

G Google 45 f Facebook 32 🍎 Apple 28 M Microsoft 25

The key insight is to normalize emails by removing dots from local names and ignoring text after plus signs, then use a hash set to count unique addresses. Best approach uses a hash set for O(1) uniqueness checking. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force Comparison

⏱️ Time: O(n²) Space: O(n)

For each email, normalize it and compare with all previously processed normalized emails to check if it's unique. This requires nested loops and doesn't use any efficient data structure.

Hash Set for Unique Tracking

⏱️ Time: O(n) Space: O(n)

Normalize each email by removing dots from local part and handling plus signs, then add to a hash set. The set automatically handles uniqueness, so we just return its size.

Brute Force Comparison — Algorithm Steps

Normalize each email by removing dots and handling plus signs
Compare each normalized email with all previous ones
Count unique addresses

Visualization

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Step-by-Step Walkthrough

Normalize Email

Remove dots and handle plus signs in local part

Compare

Check if normalized email exists in previous results

Count Unique

Add to list if unique, return final count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char* normalize(char* email) {
    char* result = malloc(strlen(email) + 1);
    char* atPos = strchr(email, '@');
    int localLen = atPos - email;
    
    char local[localLen + 1];
    strncpy(local, email, localLen);
    local[localLen] = '\0';
    
    char* plusPos = strchr(local, '+');
    if (plusPos) {
        *plusPos = '\0';
    }
    
    char normalizedLocal[strlen(local) + 1];
    int j = 0;
    for (int i = 0; local[i]; i++) {
        if (local[i] != '.') {
            normalizedLocal[j++] = local[i];
        }
    }
    normalizedLocal[j] = '\0';
    
    sprintf(result, "%s%s", normalizedLocal, atPos);
    return result;
}

int solution(char emails[][100], int emailCount) {
    char normalized[1000][100];
    int count = 0;
    
    for (int i = 0; i < emailCount; i++) {
        char* norm = normalize(emails[i]);
        int found = 0;
        
        for (int j = 0; j < count; j++) {
            if (strcmp(normalized[j], norm) == 0) {
                found = 1;
                break;
            }
        }
        
        if (!found) {
            strcpy(normalized[count], norm);
            count++;
        }
        
        free(norm);
    }
    
    return count;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    char emails[1000][100];
    int emailCount = 0;
    
    char* token = strtok(line + 1, ",]");
    while (token) {
        while (*token == ' ' || *token == '"') token++;
        char* end = token + strlen(token) - 1;
        while (end > token && (*end == ' ' || *end == '"' || *end == '\n')) end--;
        *(end + 1) = '\0';
        
        strcpy(emails[emailCount], token);
        emailCount++;
        
        token = strtok(NULL, ",]");
    }
    
    int result = solution(emails, emailCount);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops comparing each email with all previous emails

✓ Linear Growth

Space Complexity

O(n)

Store normalized emails for comparison

⚡ Linearithmic Space

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Common Approaches

Brute Force Comparison — Algorithm Steps

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Code -

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