721. Accounts Merge - Explanation

Description

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 = [
    ["neet","neet@gmail.com","neet_dsa@gmail.com"],
    ["alice","alice@gmail.com"],
    ["neet","bob@gmail.com","neet@gmail.com"],
    ["neet","neetcode@gmail.com"]
]

Output: [["neet","bob@gmail.com","neet@gmail.com","neet_dsa@gmail.com"],["alice","alice@gmail.com"],["neet","neetcode@gmail.com"]]

Example 2:

Input: accounts = [
    ["James","james@mail.com"],
    ["James","james@mail.co"]
]

Output: [["James","james@mail.com"],["James","james@mail.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.

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Hash Maps - Using dictionaries to map keys to values and track relationships
Graph Representation - Building adjacency lists from problem constraints
DFS/BFS Traversal - Exploring all nodes in a connected component
Union-Find (Disjoint Set Union) - Understanding how to efficiently merge and query sets (for optimal solution)
Connected Components - Recognizing when a problem is about grouping related elements

1. Depth First Search

Intuition

This is a graph connectivity problem in disguise. If two accounts share an email, they belong to the same person and should be merged. We can model this as a graph where emails are nodes, and emails within the same account are connected by edges. Finding all emails belonging to one person becomes finding all nodes in a connected component. dfs naturally explores an entire component, collecting all connected emails.

Algorithm

Assign a unique index to each email and track which account it first appeared in.
Build an adjacency list connecting consecutive emails within each account.
For each unvisited email, run dfs to collect all emails in that connected component.
Group the collected emails by the account index of the starting email.
For each group, sort the emails and prepend the account name.
Return the merged accounts.

class Solution:
    def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
        n = len(accounts)
        emailIdx = {} # email -> id
        emails = [] # set of emails of all accounts
        emailToAcc = {} # email_index -> account_Id

        m = 0
        for accId, a in enumerate(accounts):
            for i in range(1, len(a)):
                email = a[i]
                if email in emailIdx:
                    continue
                emails.append(email)
                emailIdx[email] = m
                emailToAcc[m] = accId
                m += 1

        adj = [[] for _ in range(m)]
        for a in accounts:
            for i in range(2, len(a)):
                id1 = emailIdx[a[i]]
                id2 = emailIdx[a[i - 1]]
                adj[id1].append(id2)
                adj[id2].append(id1)

        emailGroup = defaultdict(list) # index of acc -> list of emails
        visited = [False] * m
        def dfs(node, accId):
            visited[node] = True
            emailGroup[accId].append(emails[node])
            for nei in adj[node]:
                if not visited[nei]:
                    dfs(nei, accId)

        for i in range(m):
            if not visited[i]:
                dfs(i, emailToAcc[i])

        res = []
        for accId in emailGroup:
            name = accounts[accId][0]
            res.append([name] + sorted(emailGroup[accId]))

        return res

public class Solution {
    private Map<String, Integer> emailIdx = new HashMap<>(); // email -> id
    private List<String> emails = new ArrayList<>(); // set of emails of all accounts
    private Map<Integer, Integer> emailToAcc = new HashMap<>(); // email_index -> account_Id
    private List<List<Integer>> adj;
    private Map<Integer, List<String>> emailGroup = new HashMap<>(); // index of acc -> list of emails
    private boolean[] visited;

    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        int n = accounts.size();
        int m = 0;

        // Build email index and mappings
        for (int accId = 0; accId < n; accId++) {
            List<String> account = accounts.get(accId);
            for (int i = 1; i < account.size(); i++) {
                String email = account.get(i);
                if (!emailIdx.containsKey(email)) {
                    emails.add(email);
                    emailIdx.put(email, m);
                    emailToAcc.put(m, accId);
                    m++;
                }
            }
        }

        // Build adjacency list
        adj = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            adj.add(new ArrayList<>());
        }
        for (List<String> account : accounts) {
            for (int i = 2; i < account.size(); i++) {
                int id1 = emailIdx.get(account.get(i));
                int id2 = emailIdx.get(account.get(i - 1));
                adj.get(id1).add(id2);
                adj.get(id2).add(id1);
            }
        }

        // Initialize visited array
        visited = new boolean[m];

        // DFS traversal
        for (int i = 0; i < m; i++) {
            if (!visited[i]) {
                int accId = emailToAcc.get(i);
                emailGroup.putIfAbsent(accId, new ArrayList<>());
                dfs(i, accId);
            }
        }

        // Build result
        List<List<String>> res = new ArrayList<>();
        for (int accId : emailGroup.keySet()) {
            List<String> group = emailGroup.get(accId);
            Collections.sort(group);
            List<String> merged = new ArrayList<>();
            merged.add(accounts.get(accId).get(0));
            merged.addAll(group);
            res.add(merged);
        }

        return res;
    }

    private void dfs(int node, int accId) {
        visited[node] = true;
        emailGroup.get(accId).add(emails.get(node));
        for (int neighbor : adj.get(node)) {
            if (!visited[neighbor]) {
                dfs(neighbor, accId);
            }
        }
    }
}

class Solution {
    unordered_map<string, int> emailIdx; // email -> id
    vector<string> emails; // set of emails of all accounts
    unordered_map<int, int> emailToAcc; // email_index -> account_Id
    vector<vector<int>> adj;
    unordered_map<int, vector<string>> emailGroup; // index of acc -> list of emails
    vector<bool> visited;

public:
    vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
        int n = accounts.size();
        int m = 0;

        // Build email index and mappings
        for (int accId = 0; accId < n; accId++) {
            vector<string>& account = accounts[accId];
            for (int i = 1; i < account.size(); i++) {
                string& email = account[i];
                if (emailIdx.find(email) == emailIdx.end()) {
                    emails.push_back(email);
                    emailIdx[email] = m;
                    emailToAcc[m] = accId;
                    m++;
                }
            }
        }

        // Build adjacency list
        adj.resize(m);
        for (auto& account : accounts) {
            for (int i = 2; i < account.size(); i++) {
                int id1 = emailIdx[account[i]];
                int id2 = emailIdx[account[i - 1]];
                adj[id1].push_back(id2);
                adj[id2].push_back(id1);
            }
        }

        visited.resize(m, false);
        // DFS traversal
        for (int i = 0; i < m; i++) {
            if (!visited[i]) {
                int accId = emailToAcc[i];
                dfs(i, accId);
            }
        }

        // Build result
        vector<vector<string>> res;
        for (auto& [accId, group] : emailGroup) {
            sort(group.begin(), group.end());
            vector<string> merged;
            merged.push_back(accounts[accId][0]);
            merged.insert(merged.end(), group.begin(), group.end());
            res.push_back(merged);
        }

        return res;
    }

private:
    void dfs(int node, int& accId) {
        visited[node] = true;
        emailGroup[accId].push_back(emails[node]);
        for (int& neighbor : adj[node]) {
            if (!visited[neighbor]) {
                dfs(neighbor, accId);
            }
        }
    }
};

class Solution {
    /**
     * @param {string[][]} accounts
     * @return {string[][]}
     */
    accountsMerge(accounts) {
        const emailIdx = new Map(); // email -> id
        const emails = []; // set of emails of all accounts
        const emailToAcc = new Map(); // email_index -> account_Id
        const adj = [];
        const emailGroup = new Map(); // index of acc -> list of emails
        let visited = [];

        const n = accounts.length;
        let m = 0;

        // Build email index and mappings
        for (let accId = 0; accId < n; accId++) {
            const account = accounts[accId];
            for (let i = 1; i < account.length; i++) {
                const email = account[i];
                if (!emailIdx.has(email)) {
                    emails.push(email);
                    emailIdx.set(email, m);
                    emailToAcc.set(m, accId);
                    m++;
                }
            }
        }

        // Build adjacency list
        for (let i = 0; i < m; i++) {
            adj.push([]);
        }
        for (const account of accounts) {
            for (let i = 2; i < account.length; i++) {
                const id1 = emailIdx.get(account[i]);
                const id2 = emailIdx.get(account[i - 1]);
                adj[id1].push(id2);
                adj[id2].push(id1);
            }
        }

        // Initialize visited array
        visited = Array(m).fill(false);

        // DFS traversal
        const dfs = (node, accId) => {
            visited[node] = true;
            emailGroup.get(accId).push(emails[node]);
            for (const neighbor of adj[node]) {
                if (!visited[neighbor]) {
                    dfs(neighbor, accId);
                }
            }
        };

        for (let i = 0; i < m; i++) {
            if (!visited[i]) {
                const accId = emailToAcc.get(i);
                if (!emailGroup.has(accId)) {
                    emailGroup.set(accId, []);
                }
                dfs(i, accId);
            }
        }

        // Build result
        const res = [];
        for (const [accId, group] of emailGroup.entries()) {
            group.sort();
            const merged = [accounts[accId][0], ...group];
            res.push(merged);
        }

        return res;
    }
}

public class Solution {
    private Dictionary<string, int> emailIdx = new Dictionary<string, int>();
    private List<string> emails = new List<string>();
    private List<List<int>> adj;
    private bool[] visited;
    private Dictionary<int, List<string>> components = new Dictionary<int, List<string>>();
    private Dictionary<int, string> componentName = new Dictionary<int, string>();

    public List<List<string>> AccountsMerge(List<List<string>> accounts) {
        int m = 0;

        for (int accId = 0; accId < accounts.Count; accId++) {
            var account = accounts[accId];
            for (int i = 1; i < account.Count; i++) {
                string email = account[i];
                if (!emailIdx.ContainsKey(email)) {
                    emailIdx[email] = m++;
                    emails.Add(email);
                }
            }
        }

        adj = new List<List<int>>();
        for (int i = 0; i < m; i++) adj.Add(new List<int>());

        foreach (var account in accounts) {
            for (int i = 2; i < account.Count; i++) {
                int u = emailIdx[account[i - 1]];
                int v = emailIdx[account[i]];
                adj[u].Add(v);
                adj[v].Add(u);
            }
        }

        visited = new bool[m];

        foreach (var account in accounts) {
            string name = account[0];
            foreach (var email in account.Skip(1)) {
                int idx = emailIdx[email];
                if (!visited[idx]) {
                    components[idx] = new List<string>();
                    componentName[idx] = name;
                    Dfs(idx, idx);
                }
            }
        }

        var res = new List<List<string>>();
        foreach (var kvp in components) {
            var group = kvp.Value;
            group.Sort(StringComparer.Ordinal);
            var merged = new List<string> { componentName[kvp.Key] };
            merged.AddRange(group);
            res.Add(merged);
        }

        return res;
    }

    private void Dfs(int node, int root) {
        visited[node] = true;
        components[root].Add(emails[node]);
        foreach (int nei in adj[node]) {
            if (!visited[nei]) {
                Dfs(nei, root);
            }
        }
    }
}

func accountsMerge(accounts [][]string) [][]string {
    emailIdx := make(map[string]int)
    emails := []string{}
    emailToAcc := make(map[int]int)

    m := 0
    for accId, account := range accounts {
        for i := 1; i < len(account); i++ {
            email := account[i]
            if _, exists := emailIdx[email]; !exists {
                emails = append(emails, email)
                emailIdx[email] = m
                emailToAcc[m] = accId
                m++
            }
        }
    }

    adj := make([][]int, m)
    for i := range adj {
        adj[i] = []int{}
    }
    for _, account := range accounts {
        for i := 2; i < len(account); i++ {
            id1 := emailIdx[account[i]]
            id2 := emailIdx[account[i-1]]
            adj[id1] = append(adj[id1], id2)
            adj[id2] = append(adj[id2], id1)
        }
    }

    emailGroup := make(map[int][]string)
    visited := make([]bool, m)

    var dfs func(node, accId int)
    dfs = func(node, accId int) {
        visited[node] = true
        emailGroup[accId] = append(emailGroup[accId], emails[node])
        for _, nei := range adj[node] {
            if !visited[nei] {
                dfs(nei, accId)
            }
        }
    }

    for i := 0; i < m; i++ {
        if !visited[i] {
            dfs(i, emailToAcc[i])
        }
    }

    res := [][]string{}
    for accId, group := range emailGroup {
        name := accounts[accId][0]
        sort.Strings(group)
        merged := append([]string{name}, group...)
        res = append(res, merged)
    }

    return res
}

class Solution {
    private lateinit var emailIdx: MutableMap<String, Int>
    private lateinit var emails: MutableList<String>
    private lateinit var adj: MutableList<MutableList<Int>>
    private lateinit var emailGroup: MutableMap<Int, MutableList<String>>
    private lateinit var visited: BooleanArray

    fun accountsMerge(accounts: List<List<String>>): List<List<String>> {
        emailIdx = mutableMapOf()
        emails = mutableListOf()
        val emailToAcc = mutableMapOf<Int, Int>()

        var m = 0
        for ((accId, account) in accounts.withIndex()) {
            for (i in 1 until account.size) {
                val email = account[i]
                if (email !in emailIdx) {
                    emails.add(email)
                    emailIdx[email] = m
                    emailToAcc[m] = accId
                    m++
                }
            }
        }

        adj = MutableList(m) { mutableListOf() }
        for (account in accounts) {
            for (i in 2 until account.size) {
                val id1 = emailIdx[account[i]]!!
                val id2 = emailIdx[account[i - 1]]!!
                adj[id1].add(id2)
                adj[id2].add(id1)
            }
        }

        emailGroup = mutableMapOf()
        visited = BooleanArray(m)

        for (i in 0 until m) {
            if (!visited[i]) {
                emailGroup[emailToAcc[i]!!] = mutableListOf()
                dfs(i, emailToAcc[i]!!)
            }
        }

        val res = mutableListOf<List<String>>()
        for ((accId, group) in emailGroup) {
            val name = accounts[accId][0]
            group.sort()
            res.add(listOf(name) + group)
        }

        return res
    }

    private fun dfs(node: Int, accId: Int) {
        visited[node] = true
        emailGroup[accId]!!.add(emails[node])
        for (nei in adj[node]) {
            if (!visited[nei]) {
                dfs(nei, accId)
            }
        }
    }
}

class Solution {
    private var emailIdx = [String: Int]()
    private var emails = [String]()
    private var adj = [[Int]]()
    private var emailGroup = [Int: [String]]()
    private var visited = [Bool]()

    func accountsMerge(_ accounts: [[String]]) -> [[String]] {
        emailIdx = [:]
        emails = []
        var emailToAcc = [Int: Int]()

        var m = 0
        for (accId, account) in accounts.enumerated() {
            for i in 1..<account.count {
                let email = account[i]
                if emailIdx[email] == nil {
                    emails.append(email)
                    emailIdx[email] = m
                    emailToAcc[m] = accId
                    m += 1
                }
            }
        }

        adj = Array(repeating: [Int](), count: m)
        for account in accounts {
            for i in 2..<account.count {
                let id1 = emailIdx[account[i]]!
                let id2 = emailIdx[account[i - 1]]!
                adj[id1].append(id2)
                adj[id2].append(id1)
            }
        }

        emailGroup = [:]
        visited = Array(repeating: false, count: m)

        for i in 0..<m {
            if !visited[i] {
                emailGroup[emailToAcc[i]!] = []
                dfs(i, emailToAcc[i]!)
            }
        }

        var res = [[String]]()
        for (accId, group) in emailGroup {
            let name = accounts[accId][0]
            let sorted = group.sorted()
            res.append([name] + sorted)
        }

        return res
    }

    private func dfs(_ node: Int, _ accId: Int) {
        visited[node] = true
        emailGroup[accId]!.append(emails[node])
        for nei in adj[node] {
            if !visited[nei] {
                dfs(nei, accId)
            }
        }
    }
}

impl Solution {
    pub fn accounts_merge(accounts: Vec<Vec<String>>) -> Vec<Vec<String>> {
        let n = accounts.len();
        let mut email_idx: HashMap<String, usize> = HashMap::new();
        let mut emails: Vec<String> = Vec::new();
        let mut email_to_acc: HashMap<usize, usize> = HashMap::new();

        let mut m = 0;
        for (acc_id, account) in accounts.iter().enumerate() {
            for i in 1..account.len() {
                let email = &account[i];
                if !email_idx.contains_key(email) {
                    emails.push(email.clone());
                    email_idx.insert(email.clone(), m);
                    email_to_acc.insert(m, acc_id);
                    m += 1;
                }
            }
        }

        let mut adj = vec![vec![]; m];
        for account in &accounts {
            for i in 2..account.len() {
                let id1 = email_idx[&account[i]];
                let id2 = email_idx[&account[i - 1]];
                adj[id1].push(id2);
                adj[id2].push(id1);
            }
        }

        let mut email_group: HashMap<usize, Vec<String>> = HashMap::new();
        let mut visited = vec![false; m];

        fn dfs(
            node: usize, acc_id: usize, adj: &Vec<Vec<usize>>,
            emails: &Vec<String>, visited: &mut Vec<bool>,
            email_group: &mut HashMap<usize, Vec<String>>,
        ) {
            visited[node] = true;
            email_group.entry(acc_id).or_default().push(emails[node].clone());
            for &nei in &adj[node] {
                if !visited[nei] {
                    dfs(nei, acc_id, adj, emails, visited, email_group);
                }
            }
        }

        for i in 0..m {
            if !visited[i] {
                dfs(i, email_to_acc[&i], &adj, &emails, &mut visited, &mut email_group);
            }
        }

        let mut res = Vec::new();
        for (acc_id, mut group) in email_group {
            let name = accounts[acc_id][0].clone();
            group.sort();
            let mut merged = vec![name];
            merged.extend(group);
            res.push(merged);
        }

        res
    }
}

Time & Space Complexity

Time complexity: $O((n * m)\log (n * m))$
Space complexity: $O(n * m)$

Where $n$ is the number of accounts and $m$ is the number of emails.

2. Breadth First Search

Intuition

bfs provides an alternative way to explore connected components. Starting from any unvisited email, we use a queue to visit all reachable emails level by level. Each email we dequeue gets added to the current component, and its unvisited neighbors are enqueued. The result is the same as dfs, but bfs uses iteration with a queue instead of recursion.

Algorithm

Assign a unique index to each email and track which account it first appeared in.
Build an adjacency list connecting consecutive emails within each account.
For each unvisited email, start bfs:
- Initialize a queue with the starting email and mark it visited.
- While the queue is not empty, dequeue an email, add it to the current group, and enqueue its unvisited neighbors.
Group the collected emails by the account index of the starting email.
For each group, sort the emails and prepend the account name.
Return the merged accounts.

class Solution:
    def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
        n = len(accounts)
        emailIdx = {} # email -> id
        emails = [] # set of emails of all accounts
        emailToAcc = {} # email_index -> account_Id

        m = 0
        for accId, a in enumerate(accounts):
            for i in range(1, len(a)):
                email = a[i]
                if email in emailIdx:
                    continue
                emails.append(email)
                emailIdx[email] = m
                emailToAcc[m] = accId
                m += 1

        adj = [[] for _ in range(m)]
        for a in accounts:
            for i in range(2, len(a)):
                id1 = emailIdx[a[i]]
                id2 = emailIdx[a[i - 1]]
                adj[id1].append(id2)
                adj[id2].append(id1)

        emailGroup = defaultdict(list) # index of acc -> list of emails
        visited = [False] * m

        def bfs(start, accId):
            queue = deque([start])
            visited[start] = True
            while queue:
                node = queue.popleft()
                emailGroup[accId].append(emails[node])
                for nei in adj[node]:
                    if not visited[nei]:
                        visited[nei] = True
                        queue.append(nei)

        for i in range(m):
            if not visited[i]:
                bfs(i, emailToAcc[i])

        res = []
        for accId in emailGroup:
            name = accounts[accId][0]
            res.append([name] + sorted(emailGroup[accId]))

        return res

public class Solution {
    private Map<String, Integer> emailIdx = new HashMap<>(); // email -> id
    private List<String> emails = new ArrayList<>(); // set of emails of all accounts
    private Map<Integer, Integer> emailToAcc = new HashMap<>(); // email_index -> account_Id
    private List<List<Integer>> adj;
    private Map<Integer, List<String>> emailGroup = new HashMap<>(); // index of acc -> list of emails
    private boolean[] visited;

    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        int n = accounts.size();
        int m = 0;

        // Build email index and mappings
        for (int accId = 0; accId < n; accId++) {
            List<String> account = accounts.get(accId);
            for (int i = 1; i < account.size(); i++) {
                String email = account.get(i);
                if (!emailIdx.containsKey(email)) {
                    emails.add(email);
                    emailIdx.put(email, m);
                    emailToAcc.put(m, accId);
                    m++;
                }
            }
        }

        // Build adjacency list
        adj = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            adj.add(new ArrayList<>());
        }
        for (List<String> account : accounts) {
            for (int i = 2; i < account.size(); i++) {
                int id1 = emailIdx.get(account.get(i));
                int id2 = emailIdx.get(account.get(i - 1));
                adj.get(id1).add(id2);
                adj.get(id2).add(id1);
            }
        }

        // Initialize visited array
        visited = new boolean[m];

        // BFS traversal
        for (int i = 0; i < m; i++) {
            if (!visited[i]) {
                int accId = emailToAcc.get(i);
                emailGroup.putIfAbsent(accId, new ArrayList<>());
                bfs(i, accId);
            }
        }

        // Build result
        List<List<String>> res = new ArrayList<>();
        for (int accId : emailGroup.keySet()) {
            List<String> group = emailGroup.get(accId);
            Collections.sort(group);
            List<String> merged = new ArrayList<>();
            merged.add(accounts.get(accId).get(0));
            merged.addAll(group);
            res.add(merged);
        }

        return res;
    }

    private void bfs(int start, int accId) {
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(start);
        visited[start] = true;

        while (!queue.isEmpty()) {
            int node = queue.poll();
            emailGroup.get(accId).add(emails.get(node));
            for (int neighbor : adj.get(node)) {
                if (!visited[neighbor]) {
                    visited[neighbor] = true;
                    queue.offer(neighbor);
                }
            }
        }
    }
}

class Solution {
    unordered_map<string, int> emailIdx; // email -> id
    vector<string> emails; // set of emails of all accounts
    unordered_map<int, int> emailToAcc; // email_index -> account_Id
    vector<vector<int>> adj;
    unordered_map<int, vector<string>> emailGroup; // index of acc -> list of emails
    vector<bool> visited;

public:
    vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
        int n = accounts.size();
        int m = 0;

        // Build email index and mappings
        for (int accId = 0; accId < n; accId++) {
            vector<string>& account = accounts[accId];
            for (int i = 1; i < account.size(); i++) {
                string& email = account[i];
                if (emailIdx.find(email) == emailIdx.end()) {
                    emails.push_back(email);
                    emailIdx[email] = m;
                    emailToAcc[m] = accId;
                    m++;
                }
            }
        }

        // Build adjacency list
        adj.resize(m);
        for (auto& account : accounts) {
            for (int i = 2; i < account.size(); i++) {
                int id1 = emailIdx[account[i]];
                int id2 = emailIdx[account[i - 1]];
                adj[id1].push_back(id2);
                adj[id2].push_back(id1);
            }
        }

        visited.resize(m, false);
        // BFS traversal
        for (int i = 0; i < m; i++) {
            if (!visited[i]) {
                int accId = emailToAcc[i];
                bfs(i, accId);
            }
        }

        // Build result
        vector<vector<string>> res;
        for (auto& [accId, group] : emailGroup) {
            sort(group.begin(), group.end());
            vector<string> merged;
            merged.push_back(accounts[accId][0]);
            merged.insert(merged.end(), group.begin(), group.end());
            res.push_back(merged);
        }

        return res;
    }

private:
    void bfs(int start, int accId) {
        queue<int> q;
        q.push(start);
        visited[start] = true;

        while (!q.empty()) {
            int node = q.front();
            q.pop();
            emailGroup[accId].push_back(emails[node]);
            for (int& neighbor : adj[node]) {
                if (!visited[neighbor]) {
                    visited[neighbor] = true;
                    q.push(neighbor);
                }
            }
        }
    }
};

class Solution {
    /**
     * @param {string[][]} accounts
     * @return {string[][]}
     */
    accountsMerge(accounts) {
        const emailIdx = new Map(); // email -> id
        const emails = []; // set of emails of all accounts
        const emailToAcc = new Map(); // email_index -> account_Id
        const adj = [];
        const emailGroup = new Map(); // index of acc -> list of emails
        let visited = [];

        const n = accounts.length;
        let m = 0;

        // Build email index and mappings
        for (let accId = 0; accId < n; accId++) {
            const account = accounts[accId];
            for (let i = 1; i < account.length; i++) {
                const email = account[i];
                if (!emailIdx.has(email)) {
                    emails.push(email);
                    emailIdx.set(email, m);
                    emailToAcc.set(m, accId);
                    m++;
                }
            }
        }

        // Build adjacency list
        for (let i = 0; i < m; i++) {
            adj.push([]);
        }
        for (const account of accounts) {
            for (let i = 2; i < account.length; i++) {
                const id1 = emailIdx.get(account[i]);
                const id2 = emailIdx.get(account[i - 1]);
                adj[id1].push(id2);
                adj[id2].push(id1);
            }
        }

        // Initialize visited array
        visited = Array(m).fill(false);

        // BFS traversal
        const bfs = (start, accId) => {
            const queue = new Queue([start]);
            visited[start] = true;

            while (!queue.isEmpty()) {
                const node = queue.pop();
                emailGroup.get(accId).push(emails[node]);
                for (const neighbor of adj[node]) {
                    if (!visited[neighbor]) {
                        visited[neighbor] = true;
                        queue.push(neighbor);
                    }
                }
            }
        };

        for (let i = 0; i < m; i++) {
            if (!visited[i]) {
                const accId = emailToAcc.get(i);
                if (!emailGroup.has(accId)) {
                    emailGroup.set(accId, []);
                }
                bfs(i, accId);
            }
        }

        // Build result
        const res = [];
        for (const [accId, group] of emailGroup.entries()) {
            group.sort();
            const merged = [accounts[accId][0], ...group];
            res.push(merged);
        }

        return res;
    }
}

public class Solution {
    public List<List<string>> AccountsMerge(List<List<string>> accounts) {
        int n = accounts.Count;
        Dictionary<string, int> emailIdx = new Dictionary<string, int>();
        List<string> emails = new List<string>();
        Dictionary<int, int> emailToAcc = new Dictionary<int, int>();

        int m = 0;
        for (int accId = 0; accId < n; accId++) {
            var account = accounts[accId];
            for (int i = 1; i < account.Count; i++) {
                string email = account[i];
                if (!emailIdx.ContainsKey(email)) {
                    emailIdx[email] = m;
                    emails.Add(email);
                    emailToAcc[m] = accId;
                    m++;
                }
            }
        }

        List<List<int>> adj = new List<List<int>>();
        for (int i = 0; i < m; i++) adj.Add(new List<int>());

        foreach (var account in accounts) {
            for (int i = 2; i < account.Count; i++) {
                int id1 = emailIdx[account[i]];
                int id2 = emailIdx[account[i - 1]];
                adj[id1].Add(id2);
                adj[id2].Add(id1);
            }
        }

        Dictionary<int, List<string>> emailGroup = new Dictionary<int, List<string>>();
        bool[] visited = new bool[m];

        void Bfs(int start, int accId) {
            Queue<int> queue = new Queue<int>();
            queue.Enqueue(start);
            visited[start] = true;

            if (!emailGroup.ContainsKey(accId))
                emailGroup[accId] = new List<string>();

            while (queue.Count > 0) {
                int node = queue.Dequeue();
                emailGroup[accId].Add(emails[node]);

                foreach (int nei in adj[node]) {
                    if (!visited[nei]) {
                        visited[nei] = true;
                        queue.Enqueue(nei);
                    }
                }
            }
        }

        for (int i = 0; i < m; i++) {
            if (!visited[i]) {
                Bfs(i, emailToAcc[i]);
            }
        }

        List<List<string>> res = new List<List<string>>();
        foreach (var kvp in emailGroup) {
            int accId = kvp.Key;
            string name = accounts[accId][0];
            List<string> merged = new List<string> { name };
            kvp.Value.Sort(StringComparer.Ordinal);
            merged.AddRange(kvp.Value);
            res.Add(merged);
        }

        return res;
    }
}

func accountsMerge(accounts [][]string) [][]string {
    emailIdx := make(map[string]int)
    emails := []string{}
    emailToAcc := make(map[int]int)

    m := 0
    for accId, account := range accounts {
        for i := 1; i < len(account); i++ {
            email := account[i]
            if _, exists := emailIdx[email]; !exists {
                emails = append(emails, email)
                emailIdx[email] = m
                emailToAcc[m] = accId
                m++
            }
        }
    }

    adj := make([][]int, m)
    for i := range adj {
        adj[i] = []int{}
    }
    for _, account := range accounts {
        for i := 2; i < len(account); i++ {
            id1 := emailIdx[account[i]]
            id2 := emailIdx[account[i-1]]
            adj[id1] = append(adj[id1], id2)
            adj[id2] = append(adj[id2], id1)
        }
    }

    emailGroup := make(map[int][]string)
    visited := make([]bool, m)

    bfs := func(start, accId int) {
        queue := []int{start}
        visited[start] = true
        for len(queue) > 0 {
            node := queue[0]
            queue = queue[1:]
            emailGroup[accId] = append(emailGroup[accId], emails[node])
            for _, nei := range adj[node] {
                if !visited[nei] {
                    visited[nei] = true
                    queue = append(queue, nei)
                }
            }
        }
    }

    for i := 0; i < m; i++ {
        if !visited[i] {
            bfs(i, emailToAcc[i])
        }
    }

    res := [][]string{}
    for accId, group := range emailGroup {
        name := accounts[accId][0]
        sort.Strings(group)
        merged := append([]string{name}, group...)
        res = append(res, merged)
    }

    return res
}

class Solution {
    fun accountsMerge(accounts: List<List<String>>): List<List<String>> {
        val emailIdx = mutableMapOf<String, Int>()
        val emails = mutableListOf<String>()
        val emailToAcc = mutableMapOf<Int, Int>()

        var m = 0
        for ((accId, account) in accounts.withIndex()) {
            for (i in 1 until account.size) {
                val email = account[i]
                if (email !in emailIdx) {
                    emails.add(email)
                    emailIdx[email] = m
                    emailToAcc[m] = accId
                    m++
                }
            }
        }

        val adj = MutableList(m) { mutableListOf<Int>() }
        for (account in accounts) {
            for (i in 2 until account.size) {
                val id1 = emailIdx[account[i]]!!
                val id2 = emailIdx[account[i - 1]]!!
                adj[id1].add(id2)
                adj[id2].add(id1)
            }
        }

        val emailGroup = mutableMapOf<Int, MutableList<String>>()
        val visited = BooleanArray(m)

        fun bfs(start: Int, accId: Int) {
            val queue = ArrayDeque<Int>()
            queue.add(start)
            visited[start] = true
            emailGroup[accId] = mutableListOf()

            while (queue.isNotEmpty()) {
                val node = queue.removeFirst()
                emailGroup[accId]!!.add(emails[node])
                for (nei in adj[node]) {
                    if (!visited[nei]) {
                        visited[nei] = true
                        queue.add(nei)
                    }
                }
            }
        }

        for (i in 0 until m) {
            if (!visited[i]) {
                bfs(i, emailToAcc[i]!!)
            }
        }

        val res = mutableListOf<List<String>>()
        for ((accId, group) in emailGroup) {
            val name = accounts[accId][0]
            group.sort()
            res.add(listOf(name) + group)
        }

        return res
    }
}

class Solution {
    func accountsMerge(_ accounts: [[String]]) -> [[String]] {
        var emailIdx = [String: Int]()
        var emails = [String]()
        var emailToAcc = [Int: Int]()

        var m = 0
        for (accId, account) in accounts.enumerated() {
            for i in 1..<account.count {
                let email = account[i]
                if emailIdx[email] == nil {
                    emails.append(email)
                    emailIdx[email] = m
                    emailToAcc[m] = accId
                    m += 1
                }
            }
        }

        var adj = Array(repeating: [Int](), count: m)
        for account in accounts {
            for i in 2..<account.count {
                let id1 = emailIdx[account[i]]!
                let id2 = emailIdx[account[i - 1]]!
                adj[id1].append(id2)
                adj[id2].append(id1)
            }
        }

        var emailGroup = [Int: [String]]()
        var visited = Array(repeating: false, count: m)

        func bfs(_ start: Int, _ accId: Int) {
            var queue = [start]
            visited[start] = true
            emailGroup[accId] = []

            while !queue.isEmpty {
                let node = queue.removeFirst()
                emailGroup[accId]!.append(emails[node])
                for nei in adj[node] {
                    if !visited[nei] {
                        visited[nei] = true
                        queue.append(nei)
                    }
                }
            }
        }

        for i in 0..<m {
            if !visited[i] {
                bfs(i, emailToAcc[i]!)
            }
        }

        var res = [[String]]()
        for (accId, group) in emailGroup {
            let name = accounts[accId][0]
            let sorted = group.sorted()
            res.append([name] + sorted)
        }

        return res
    }
}

impl Solution {
    pub fn accounts_merge(accounts: Vec<Vec<String>>) -> Vec<Vec<String>> {
        let n = accounts.len();
        let mut email_idx: HashMap<String, usize> = HashMap::new();
        let mut emails: Vec<String> = Vec::new();
        let mut email_to_acc: HashMap<usize, usize> = HashMap::new();

        let mut m = 0;
        for (acc_id, account) in accounts.iter().enumerate() {
            for i in 1..account.len() {
                let email = &account[i];
                if !email_idx.contains_key(email) {
                    emails.push(email.clone());
                    email_idx.insert(email.clone(), m);
                    email_to_acc.insert(m, acc_id);
                    m += 1;
                }
            }
        }

        let mut adj = vec![vec![]; m];
        for account in &accounts {
            for i in 2..account.len() {
                let id1 = email_idx[&account[i]];
                let id2 = email_idx[&account[i - 1]];
                adj[id1].push(id2);
                adj[id2].push(id1);
            }
        }

        let mut email_group: HashMap<usize, Vec<String>> = HashMap::new();
        let mut visited = vec![false; m];

        for i in 0..m {
            if !visited[i] {
                let acc_id = email_to_acc[&i];
                let mut queue = VecDeque::new();
                queue.push_back(i);
                visited[i] = true;
                email_group.entry(acc_id).or_default();

                while let Some(node) = queue.pop_front() {
                    email_group.get_mut(&acc_id).unwrap().push(emails[node].clone());
                    for &nei in &adj[node] {
                        if !visited[nei] {
                            visited[nei] = true;
                            queue.push_back(nei);
                        }
                    }
                }
            }
        }

        let mut res = Vec::new();
        for (acc_id, mut group) in email_group {
            let name = accounts[acc_id][0].clone();
            group.sort();
            let mut merged = vec![name];
            merged.extend(group);
            res.push(merged);
        }

        res
    }
}

Time & Space Complexity

Time complexity: $O((n * m)\log (n * m))$
Space complexity: $O(n * m)$

Where $n$ is the number of accounts and $m$ is the number of emails.

3. Disjoint Set Union

Intuition

Union-Find (Disjoint Set Union) is designed for exactly this type of problem: grouping elements into disjoint sets and merging sets efficiently. Instead of building a graph and traversing it, we assign each account an ID and union accounts that share an email. When we see an email for the first time, we record which account it belongs to. If we see it again, we union the current account with the one that first owned it. After processing all accounts, we group emails by their account's root representative.

Algorithm

Initialize Union-Find with one node per account.
Create a map from email to the first account index that contains it.
For each account's emails:
- If the email was seen before, union the current account with the previous owner.
- Otherwise, record the current account as the owner.
For each email, find the root of its owning account and group emails by root.
For each group, sort the emails and prepend the account name.
Return the merged accounts.

class UnionFind:
    def __init__(self, n):
        self.par = [i for i in range(n)]
        self.rank = [1] * n

    def find(self, x):
        while x != self.par[x]:
            self.par[x] = self.par[self.par[x]]
            x = self.par[x]
        return x

    def union(self, x1, x2):
        p1, p2 = self.find(x1), self.find(x2)
        if p1 == p2:
            return False
        if self.rank[p1] > self.rank[p2]:
            self.par[p2] = p1
            self.rank[p1] += self.rank[p2]
        else:
            self.par[p1] = p2
            self.rank[p2] += self.rank[p1]
        return True

class Solution:
    def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
        uf = UnionFind(len(accounts))
        emailToAcc = {}  # email -> index of acc

        for i, a in enumerate(accounts):
            for e in a[1:]:
                if e in emailToAcc:
                    uf.union(i, emailToAcc[e])
                else:
                    emailToAcc[e] = i

        emailGroup = defaultdict(list)  # index of acc -> list of emails
        for e, i in emailToAcc.items():
            leader = uf.find(i)
            emailGroup[leader].append(e)

        res = []
        for i, emails in emailGroup.items():
            name = accounts[i][0]
            res.append([name] + sorted(emailGroup[i]))
        return res

class UnionFind {
    private int[] parent;
    private int[] rank;

    public UnionFind(int n) {
        parent = new int[n];
        rank = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            rank[i] = 1;
        }
    }

    public int find(int x) {
        if (x != parent[x]) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }

    public boolean union(int x1, int x2) {
        int p1 = find(x1);
        int p2 = find(x2);
        if (p1 == p2) {
            return false;
        }
        if (rank[p1] > rank[p2]) {
            parent[p2] = p1;
            rank[p1] += rank[p2];
        } else {
            parent[p1] = p2;
            rank[p2] += rank[p1];
        }
        return true;
    }
}

public class Solution {
    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        int n = accounts.size();
        UnionFind uf = new UnionFind(n);
        Map<String, Integer> emailToAcc = new HashMap<>(); // email -> index of acc

        // Build union-find structure
        for (int i = 0; i < n; i++) {
            List<String> account = accounts.get(i);
            for (int j = 1; j < account.size(); j++) {
                String email = account.get(j);
                if (emailToAcc.containsKey(email)) {
                    uf.union(i, emailToAcc.get(email));
                } else {
                    emailToAcc.put(email, i);
                }
            }
        }

        // Group emails by leader account
        Map<Integer, List<String>> emailGroup = new HashMap<>(); // index of acc -> list of emails
        for (Map.Entry<String, Integer> entry : emailToAcc.entrySet()) {
            String email = entry.getKey();
            int accId = entry.getValue();
            int leader = uf.find(accId);
            emailGroup.putIfAbsent(leader, new ArrayList<>());
            emailGroup.get(leader).add(email);
        }

        // Build result
        List<List<String>> res = new ArrayList<>();
        for (Map.Entry<Integer, List<String>> entry : emailGroup.entrySet()) {
            int accId = entry.getKey();
            List<String> emails = entry.getValue();
            Collections.sort(emails);
            List<String> merged = new ArrayList<>();
            merged.add(accounts.get(accId).get(0)); // Add account name
            merged.addAll(emails);
            res.add(merged);
        }

        return res;
    }
}

class UnionFind {
    vector<int> parent;
    vector<int> rank;

public:
    UnionFind(int n) {
        parent.resize(n);
        rank.resize(n, 1);
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
    }

    int find(int x) {
        if (x != parent[x]) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }

    bool unionSets(int x1, int x2) {
        int p1 = find(x1);
        int p2 = find(x2);
        if (p1 == p2) {
            return false;
        }
        if (rank[p1] > rank[p2]) {
            parent[p2] = p1;
            rank[p1] += rank[p2];
        } else {
            parent[p1] = p2;
            rank[p2] += rank[p1];
        }
        return true;
    }
};

class Solution {
public:
    vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
        int n = accounts.size();
        UnionFind uf(n);
        unordered_map<string, int> emailToAcc; // email -> index of acc

        // Build union-find structure
        for (int i = 0; i < n; i++) {
            for (int j = 1; j < accounts[i].size(); j++) {
                const string& email = accounts[i][j];
                if (emailToAcc.count(email)) {
                    uf.unionSets(i, emailToAcc[email]);
                } else {
                    emailToAcc[email] = i;
                }
            }
        }

        // Group emails by leader account
        map<int, vector<string>> emailGroup; // index of acc -> list of emails
        for (const auto& [email, accId] : emailToAcc) {
            int leader = uf.find(accId);
            emailGroup[leader].push_back(email);
        }

        // Build result
        vector<vector<string>> res;
        for (auto& [accId, emails] : emailGroup) {
            sort(emails.begin(), emails.end());
            vector<string> merged;
            merged.push_back(accounts[accId][0]);
            merged.insert(merged.end(), emails.begin(), emails.end());
            res.push_back(merged);
        }

        return res;
    }
};

class UnionFind {
    /**
     * @constructor
     * @param {number} n
     */
    constructor(n) {
        this.parent = Array.from({ length: n }, (_, i) => i);
        this.rank = Array(n).fill(1);
    }

    /**
     * @param {number} x
     * @return {number}
     */
    find(x) {
        if (x !== this.parent[x]) {
            this.parent[x] = this.find(this.parent[x]);
        }
        return this.parent[x];
    }

    /**
     * @param {number} x1
     * @param {number} x2
     * @return {boolean}
     */
    union(x1, x2) {
        const p1 = this.find(x1);
        const p2 = this.find(x2);
        if (p1 === p2) {
            return false;
        }
        if (this.rank[p1] > this.rank[p2]) {
            this.parent[p2] = p1;
            this.rank[p1] += this.rank[p2];
        } else {
            this.parent[p1] = p2;
            this.rank[p2] += this.rank[p1];
        }
        return true;
    }
}

class Solution {
    /**
     * @param {string[][]} accounts
     * @return {string[][]}
     */
    accountsMerge(accounts) {
        const n = accounts.length;
        const uf = new UnionFind(n);
        const emailToAcc = new Map(); // email -> index of acc

        // Build union-find structure
        for (let i = 0; i < n; i++) {
            for (let j = 1; j < accounts[i].length; j++) {
                const email = accounts[i][j];
                if (emailToAcc.has(email)) {
                    uf.union(i, emailToAcc.get(email));
                } else {
                    emailToAcc.set(email, i);
                }
            }
        }

        // Group emails by leader account
        const emailGroup = new Map(); // index of acc -> list of emails
        for (const [email, accId] of emailToAcc.entries()) {
            const leader = uf.find(accId);
            if (!emailGroup.has(leader)) {
                emailGroup.set(leader, []);
            }
            emailGroup.get(leader).push(email);
        }

        // Build result
        const res = [];
        for (const [accId, emails] of emailGroup.entries()) {
            emails.sort();
            const merged = [accounts[accId][0], ...emails];
            res.push(merged);
        }

        return res;
    }
}

public class UnionFind {
    private int[] parent;
    private int[] rank;

    public UnionFind(int n) {
        parent = new int[n];
        rank = new int[n];
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            rank[i] = 1;
        }
    }

    public int Find(int x) {
        if (x != parent[x]) {
            parent[x] = Find(parent[x]);
        }
        return parent[x];
    }

    public bool Union(int x, int y) {
        int rootX = Find(x);
        int rootY = Find(y);

        if (rootX == rootY) return false;

        if (rank[rootX] > rank[rootY]) {
            parent[rootY] = rootX;
            rank[rootX] += rank[rootY];
        } else {
            parent[rootX] = rootY;
            rank[rootY] += rank[rootX];
        }

        return true;
    }
}

public class Solution {
    public List<List<string>> AccountsMerge(List<List<string>> accounts) {
        int n = accounts.Count;
        UnionFind uf = new UnionFind(n);
        Dictionary<string, int> emailToAcc = new Dictionary<string, int>();

        for (int i = 0; i < n; i++) {
            for (int j = 1; j < accounts[i].Count; j++) {
                string email = accounts[i][j];
                if (emailToAcc.ContainsKey(email)) {
                    uf.Union(i, emailToAcc[email]);
                } else {
                    emailToAcc[email] = i;
                }
            }
        }

        Dictionary<int, List<string>> emailGroup = new Dictionary<int, List<string>>();
        foreach (var kvp in emailToAcc) {
            string email = kvp.Key;
            int leader = uf.Find(kvp.Value);
            if (!emailGroup.ContainsKey(leader)) {
                emailGroup[leader] = new List<string>();
            }
            emailGroup[leader].Add(email);
        }

        List<List<string>> res = new List<List<string>>();
        foreach (var kvp in emailGroup) {
            int accId = kvp.Key;
            List<string> emails = kvp.Value;
            emails.Sort(StringComparer.Ordinal);
            List<string> merged = new List<string> { accounts[accId][0] };
            merged.AddRange(emails);
            res.Add(merged);
        }

        return res;
    }
}

type UnionFind struct {
    parent []int
    rank   []int
}

func NewUnionFind(n int) *UnionFind {
    parent := make([]int, n)
    rank := make([]int, n)
    for i := 0; i < n; i++ {
        parent[i] = i
        rank[i] = 1
    }
    return &UnionFind{parent, rank}
}

func (uf *UnionFind) Find(x int) int {
    if x != uf.parent[x] {
        uf.parent[x] = uf.Find(uf.parent[x])
    }
    return uf.parent[x]
}

func (uf *UnionFind) Union(x, y int) bool {
    px, py := uf.Find(x), uf.Find(y)
    if px == py {
        return false
    }
    if uf.rank[px] > uf.rank[py] {
        uf.parent[py] = px
        uf.rank[px] += uf.rank[py]
    } else {
        uf.parent[px] = py
        uf.rank[py] += uf.rank[px]
    }
    return true
}

func accountsMerge(accounts [][]string) [][]string {
    n := len(accounts)
    uf := NewUnionFind(n)
    emailToAcc := make(map[string]int)

    for i, account := range accounts {
        for j := 1; j < len(account); j++ {
            email := account[j]
            if idx, exists := emailToAcc[email]; exists {
                uf.Union(i, idx)
            } else {
                emailToAcc[email] = i
            }
        }
    }

    emailGroup := make(map[int][]string)
    for email, accId := range emailToAcc {
        leader := uf.Find(accId)
        emailGroup[leader] = append(emailGroup[leader], email)
    }

    res := [][]string{}
    for accId, emails := range emailGroup {
        name := accounts[accId][0]
        sort.Strings(emails)
        merged := append([]string{name}, emails...)
        res = append(res, merged)
    }

    return res
}

class UnionFind(n: Int) {
    private val parent = IntArray(n) { it }
    private val rank = IntArray(n) { 1 }

    fun find(x: Int): Int {
        if (x != parent[x]) {
            parent[x] = find(parent[x])
        }
        return parent[x]
    }

    fun union(x: Int, y: Int): Boolean {
        val px = find(x)
        val py = find(y)
        if (px == py) return false
        if (rank[px] > rank[py]) {
            parent[py] = px
            rank[px] += rank[py]
        } else {
            parent[px] = py
            rank[py] += rank[px]
        }
        return true
    }
}

class Solution {
    fun accountsMerge(accounts: List<List<String>>): List<List<String>> {
        val n = accounts.size
        val uf = UnionFind(n)
        val emailToAcc = mutableMapOf<String, Int>()

        for (i in 0 until n) {
            for (j in 1 until accounts[i].size) {
                val email = accounts[i][j]
                if (email in emailToAcc) {
                    uf.union(i, emailToAcc[email]!!)
                } else {
                    emailToAcc[email] = i
                }
            }
        }

        val emailGroup = mutableMapOf<Int, MutableList<String>>()
        for ((email, accId) in emailToAcc) {
            val leader = uf.find(accId)
            emailGroup.getOrPut(leader) { mutableListOf() }.add(email)
        }

        val res = mutableListOf<List<String>>()
        for ((accId, emails) in emailGroup) {
            val name = accounts[accId][0]
            emails.sort()
            res.add(listOf(name) + emails)
        }

        return res
    }
}

class UnionFind {
    private var parent: [Int]
    private var rank: [Int]

    init(_ n: Int) {
        parent = Array(0..<n)
        rank = Array(repeating: 1, count: n)
    }

    func find(_ x: Int) -> Int {
        if x != parent[x] {
            parent[x] = find(parent[x])
        }
        return parent[x]
    }

    func union(_ x: Int, _ y: Int) -> Bool {
        let px = find(x)
        let py = find(y)
        if px == py { return false }
        if rank[px] > rank[py] {
            parent[py] = px
            rank[px] += rank[py]
        } else {
            parent[px] = py
            rank[py] += rank[px]
        }
        return true
    }
}

class Solution {
    func accountsMerge(_ accounts: [[String]]) -> [[String]] {
        let n = accounts.count
        let uf = UnionFind(n)
        var emailToAcc = [String: Int]()

        for i in 0..<n {
            for j in 1..<accounts[i].count {
                let email = accounts[i][j]
                if let idx = emailToAcc[email] {
                    _ = uf.union(i, idx)
                } else {
                    emailToAcc[email] = i
                }
            }
        }

        var emailGroup = [Int: [String]]()
        for (email, accId) in emailToAcc {
            let leader = uf.find(accId)
            emailGroup[leader, default: []].append(email)
        }

        var res = [[String]]()
        for (accId, emails) in emailGroup {
            let name = accounts[accId][0]
            let sorted = emails.sorted()
            res.append([name] + sorted)
        }

        return res
    }
}

struct UnionFind {
    parent: Vec<usize>,
    rank: Vec<usize>,
}

impl UnionFind {
    fn new(n: usize) -> Self {
        Self {
            parent: (0..n).collect(),
            rank: vec![1; n],
        }
    }

    fn find(&mut self, x: usize) -> usize {
        if x != self.parent[x] {
            self.parent[x] = self.find(self.parent[x]);
        }
        self.parent[x]
    }

    fn union(&mut self, x: usize, y: usize) -> bool {
        let px = self.find(x);
        let py = self.find(y);
        if px == py { return false; }
        if self.rank[px] > self.rank[py] {
            self.parent[py] = px;
            self.rank[px] += self.rank[py];
        } else {
            self.parent[px] = py;
            self.rank[py] += self.rank[px];
        }
        true
    }
}

impl Solution {
    pub fn accounts_merge(accounts: Vec<Vec<String>>) -> Vec<Vec<String>> {
        let n = accounts.len();
        let mut uf = UnionFind::new(n);
        let mut email_to_acc: HashMap<String, usize> = HashMap::new();

        for i in 0..n {
            for j in 1..accounts[i].len() {
                let email = &accounts[i][j];
                if let Some(&acc) = email_to_acc.get(email) {
                    uf.union(i, acc);
                } else {
                    email_to_acc.insert(email.clone(), i);
                }
            }
        }

        let mut email_group: HashMap<usize, Vec<String>> = HashMap::new();
        for (email, acc_id) in &email_to_acc {
            let leader = uf.find(*acc_id);
            email_group.entry(leader).or_default().push(email.clone());
        }

        let mut res = Vec::new();
        for (acc_id, mut emails) in email_group {
            let name = accounts[acc_id][0].clone();
            emails.sort();
            let mut merged = vec![name];
            merged.extend(emails);
            res.push(merged);
        }

        res
    }
}

Time & Space Complexity

Time complexity: $O((n * m)\log (n * m))$
Space complexity: $O(n * m)$

Where $n$ is the number of accounts and $m$ is the number of emails.

Common Pitfalls

Treating Same Name as Same Person

Two accounts with the same name are NOT necessarily the same person. They are only the same person if they share at least one email. Merging accounts based solely on name will produce incorrect results.

# Wrong: merging by name
if accounts[i][0] == accounts[j][0]:
    merge(i, j)
# Correct: merge only when emails overlap
if email in emailToAccount:
    union(i, emailToAccount[email])

Forgetting to Sort Emails in Output

The problem requires emails in each merged account to be sorted in lexicographical order. Forgetting this step will produce results in arbitrary order that may fail validation.

Using Wrong Account Name After Merging

When merging accounts, you must use the name from one of the original accounts in the merged group. A common mistake is losing track of which account's name to use, especially in Union-Find where you need to get the name from the representative's original account.