Accounts Merge
Given a list of `accounts` where each element `accounts[i]` is a list of strings, where the first element `accounts[i][0]` is a name, and the rest of the elements are **emails** representing emails of the account.

Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some common email to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.

After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails **in sorted order**. The accounts themselves can be returned in **any order**.

**Example 1:**

**Input:** accounts = \[\[ "John ", "johnsmith@mail.com ", "john\_newyork@mail.com "\],\[ "John ", "johnsmith@mail.com ", "john00@mail.com "\],\[ "Mary ", "mary@mail.com "\],\[ "John ", "johnnybravo@mail.com "\]\]
**Output:** \[\[ "John ", "john00@mail.com ", "john\_newyork@mail.com ", "johnsmith@mail.com "\],\[ "Mary ", "mary@mail.com "\],\[ "John ", "johnnybravo@mail.com "\]\]
**Explanation:**
The first and second John's are the same person as they have the common email  "johnsmith@mail.com ".
The third John and Mary are different people as none of their email addresses are used by other accounts.
We could return these lists in any order, for example the answer \[\['Mary', 'mary@mail.com'\], \['John', 'johnnybravo@mail.com'\], 
\['John', 'john00@mail.com', 'john\_newyork@mail.com', 'johnsmith@mail.com'\]\] would still be accepted.

**Example 2:**

**Input:** accounts = \[\[ "Gabe ", "Gabe0@m.co ", "Gabe3@m.co ", "Gabe1@m.co "\],\[ "Kevin ", "Kevin3@m.co ", "Kevin5@m.co ", "Kevin0@m.co "\],\[ "Ethan ", "Ethan5@m.co ", "Ethan4@m.co ", "Ethan0@m.co "\],\[ "Hanzo ", "Hanzo3@m.co ", "Hanzo1@m.co ", "Hanzo0@m.co "\],\[ "Fern ", "Fern5@m.co ", "Fern1@m.co ", "Fern0@m.co "\]\]
**Output:** \[\[ "Ethan ", "Ethan0@m.co ", "Ethan4@m.co ", "Ethan5@m.co "\],\[ "Gabe ", "Gabe0@m.co ", "Gabe1@m.co ", "Gabe3@m.co "\],\[ "Hanzo ", "Hanzo0@m.co ", "Hanzo1@m.co ", "Hanzo3@m.co "\],\[ "Kevin ", "Kevin0@m.co ", "Kevin3@m.co ", "Kevin5@m.co "\],\[ "Fern ", "Fern0@m.co ", "Fern1@m.co ", "Fern5@m.co "\]\]

**Constraints:**

*   `1 <= accounts.length <= 1000`
*   `2 <= accounts[i].length <= 10`
*   `1 <= accounts[i][j].length <= 30`
*   `accounts[i][0]` consists of English letters.
*   `accounts[i][j] (for j > 0)` is a valid email.

    ```java
import java.util.*;

public List<List<String>> accountsMerge(List<List<String>> accounts) {
    Map<String, Integer> emailToId = new HashMap<>();
    for (int i = 0; i < accounts.size(); ++i) {
        for (int j = 1; j < accounts.get(i).size(); ++j) {
            emailToId.put(accounts.get(i).get(j), i);
        }
    }

    List<List<String>> result = new ArrayList<>();
    boolean[] visited = new boolean[accounts.size()];
    for (int i = 0; i < accounts.size(); ++i) {
        if (!visited[i]) {
            Set<String> mergedEmails = new HashSet<>();
            dfs(i, accounts, visited, mergedEmails, emailToId);
            List<String> mergedAccount = new ArrayList<>(mergedEmails);
            Collections.sort(mergedAccount);
            mergedAccount.add(0, accounts.get(i).get(0));
            result.add(mergedAccount);
        }
    }
    return result;
}

private void dfs(int id, List<List<String>> accounts, boolean[] visited, Set<String> mergedEmails, Map<String, Integer> emailToId) {
    if (visited[id]) return;
    visited[id] = true;
    for (int j = 1; j < accounts.get(id).size(); ++j) {
        String email = accounts.get(id).get(j);
        mergedEmails.add(email);
        int nextId = emailToId.get(email);
        dfs(nextId, accounts, visited, mergedEmails, emailToId);
    }
}
```

    
    The algorithm uses a depth-first search (DFS) to merge accounts with the same email. It starts by creating a dictionary `email_to_id` to map email addresses to the account ID. Then, a visited set records visited account IDs. For each unvisited account ID, DFS is used to traverse and collect all the connected email addresses in the merged set. The result is formed by appending the sorted merged email addresses to the account's name. The final result contains all the merged accounts.
    

    ```cpp
#include <vector>
#include <string>
#include <set>
#include <map>

using namespace std;

void dfs(int id, const vector<vector<string>>& accounts, vector<bool>& visited, set<string>& merged_emails, map<string, int>& email_to_id) {
    if (visited[id]) return;
    visited[id] = true;
    for (int j = 1; j < accounts[id].size(); ++j) {
        const string& email = accounts[id][j];
        merged_emails.insert(email);
        int next_id = email_to_id[email];
        dfs(next_id, accounts, visited, merged_emails, email_to_id);
    }
}

vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
    map<string, int> email_to_id;
    for (int i = 0; i < accounts.size(); ++i)
        for (int j = 1; j < accounts[i].size(); ++j)
            email_to_id[accounts[i][j]] = i;

    vector<vector<string>> result;
    vector<bool> visited(accounts.size(), false);
    for (int i = 0; i < accounts.size(); ++i) {
        if (!visited[i]) {
            set<string> merged_emails;
            dfs(i, accounts, visited, merged_emails, email_to_id);
            vector<string> merged_account(merged_emails.begin(), merged_emails.end());
            merged_account.insert(merged_account.begin(), accounts[i][0]);
            result.push_back(merged_account);
        }
    }
    return result;
}
```

    
    The algorithm uses a depth-first search (DFS) to merge accounts with the same email. It starts by creating a dictionary `email_to_id` to map email addresses to the account ID. Then, a visited set records visited account IDs. For each unvisited account ID, DFS is used to traverse and collect all the connected email addresses in the merged set. The result is formed by appending the sorted merged email addresses to the account's name. The final result contains all the merged accounts.
    

    ```python
from collections import defaultdict

def accountsMerge(accounts):
    def dfs(id, visited, merged_emails):
        if id in visited:
            return
        visited.add(id)
        for email in accounts[id][1:]:
            merged_emails.add(email)
            dfs(email_to_id[email], visited, merged_emails)

    email_to_id = {}
    for i, account in enumerate(accounts):
        for email in account[1:]:
            email_to_id[email] = i

    result = []
    visited = set()
    for i, account in enumerate(accounts):
        if i not in visited:
            merged_emails = set()
            dfs(i, visited, merged_emails)
            result.append([account[0]] + sorted(merged_emails))
    return result
```

    
    The algorithm uses a depth-first search (DFS) to merge accounts with the same email. It starts by creating a dictionary `email_to_id` to map email addresses to the account ID. Then, a visited set records visited account IDs. For each unvisited account ID, DFS is used to traverse and collect all the connected email addresses in the merged set. The result is formed by appending the sorted merged email addresses to the account's name. The final result contains all the merged accounts.
    

    ```javascript
var accountsMerge = function(accounts) {
    function dfs(id, visited, mergedEmails) {
        if (visited.has(id)) return;
        visited.add(id);
        for (let j = 1; j < accounts[id].length; j++) {
            const email = accounts[id][j];
            mergedEmails.add(email);
            const nextId = emailToId[email];
            dfs(nextId, visited, mergedEmails);
        }
    }

    const emailToId = {};
    accounts.forEach((account, i) => {
        account.slice(1).forEach(email => {
            emailToId[email] = i;
        });
    });

    const result = [];
    const visited = new Set();
    accounts.forEach((account, i) => {
        if (!visited.has(i)) {
            const mergedEmails = new Set();
            dfs(i, visited, mergedEmails);
            result.push([account[0], ...Array.from(mergedEmails).sort()]);
        }
    });
    return result;
};
```

    
    The algorithm uses a depth-first search (DFS) to merge accounts with the same email. It starts by creating a dictionary `email_to_id` to map email addresses to the account ID. Then, a visited set records visited account IDs. For each unvisited account ID, DFS is used to traverse and collect all the connected email addresses in the merged set. The result is formed by appending the sorted merged email addresses to the account's name. The final result contains all the merged accounts.
    
from collections import defaultdict

def accountsMerge(accounts):
    def dfs(id, visited, merged_emails):
        if id in visited:
            return
        visited.add(id)
        for email in accounts[id][1:]:
            merged_emails.add(email)
            dfs(email_to_id[email], visited, merged_emails)

    email_to_id = {}
    for i, account in enumerate(accounts):
        for email in account[1:]:
            email_to_id[email] = i

    result = []
    visited = set()
    for i, account in enumerate(accounts):
        if i not in visited:
            merged_emails = set()
            dfs(i, visited, merged_emails)
            result.append([account[0]] + sorted(merged_emails))
    return result
