LeetCode 721. Accounts Merge

DFS | Union Find

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","[e-mail]","[e-mail]"],["John","[e-mail]","[e-mail]"],["Mary","[e-mail]"],["John","[e-mail]"]]
Output: [["John","[e-mail]","[e-mail]","[e-mail]"],["Mary","[e-mail]"],["John","[e-mail]"]]
Explanation:
The first and third John's are the same person as they have the common email "[e-mail]".
The second 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', '[e-mail]'], ['John', '[e-mail]'], 
['John', '[e-mail]', '[e-mail]', '[e-mail]']] would still be accepted.

Example 2:

Input: accounts = [["Gabe","[e-mail]","[e-mail]","[e-mail]"],["Kevin","[e-mail]","[e-mail]","[e-mail]"],["Ethan","[e-mail]","[e-mail]","[e-mail]"],["Hanzo","[e-mail]","[e-mail]","[e-mail]"],["Fern","[e-mail]","[e-mail]","[e-mail]"]]
Output: [["Ethan","[e-mail]","[e-mail]","[e-mail]"],["Gabe","[e-mail]","[e-mail]","[e-mail]"],["Hanzo","[e-mail]","[e-mail]","[e-mail]"],["Kevin","[e-mail]","[e-mail]","[e-mail]"],["Fern","[e-mail]","[e-mail]","[e-mail]"]]

Constraints:

• 1 <= accounts.length <= 1000

• 2 <= accounts[i].length <= 10

• 1 <= accounts[i][j] <= 30

• accounts[i][0] consists of English letters.

• accounts[i][j] (for j > 0) is a valid email.

English Version in Youtube

中文版解答Youtube Link

中文版解答Bilibili Link

Solution 1: DFS, slower when building graph

class Solution {
    void dfs(const vector<vector<int>>& graph, const vector<vector<string>>& accounts, int ind, vector<bool>& visited, set<string>& emails) {
        visited[ind] = true;
        for (int i = 1; i < accounts[ind].size(); i++) {
            emails.insert(accounts[ind][i]);
        }
        for (int neighbor : graph[ind]) {
            if (!visited[neighbor]) {
                dfs(graph, accounts, neighbor, visited, emails);
            }
        }
    }
public:
    vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
        vector<vector<string>> res;
        int n = accounts.size();
        vector<vector<int>> graph(n);
        vector<bool> visited(n, false);
        
        for (int i = 0; i < n; i++) {
            unordered_set<string> emails(accounts[i].begin()+1, accounts[i].end());
            for (int j = 0; j < i; j++) {
                if (accounts[i][0] != accounts[j][0]) {
                    continue;
                }
                for (int k = 1; k < accounts[j].size(); k++) {
                    if (emails.count(accounts[j][k]) > 0) {
                        graph[i].push_back(j);
                        graph[j].push_back(i);
                        break;
                    }
                }
            }
        }
        
        for (int i = 0; i < accounts.size(); i++) {
            if (!visited[i]) {
                set<string> emailSet;
                dfs(graph, accounts, i, visited, emailSet);
                res.resize(res.size() + 1);
                res.back().push_back(accounts[i][0]);
                for (const string& email : emailSet) {
                    res.back().push_back(email);
                }
            }
        }
        
        return res;
    }
};

Solution 2: DFS, faster when building graph

class Solution {
    void dfs(const vector<vector<int>>& graph, const vector<vector<string>>& accounts, int ind, vector<bool>& visited, set<string>& emails) {
        visited[ind] = true;
        for (int i = 1; i < accounts[ind].size(); i++) {
            emails.insert(accounts[ind][i]);
        }
        for (int neighbor : graph[ind]) {
            if (!visited[neighbor]) {
                dfs(graph, accounts, neighbor, visited, emails);
            }
        }
    }
public:
    vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
        vector<vector<string>> res;
        int n = accounts.size();
        vector<vector<int>> graph(n);
        vector<bool> visited(n, false);
        
        unordered_map<string, vector<int>> email_to_ids;
        for (int i = 0; i < n; i++) {
            for (int j = 1; j < accounts[i].size(); j++) {
                email_to_ids[accounts[i][j]].push_back(i);
            }
        }
        for (const auto& pair : email_to_ids) {
            const vector<int>& ids = pair.second;
            for (int i = 1; i < ids.size(); i++) {
                graph[ids[0]].push_back(ids[i]);
                graph[ids[i]].push_back(ids[0]);
            }
        }
        
        for (int i = 0; i < accounts.size(); i++) {
            if (!visited[i]) {
                set<string> emailSet;
                dfs(graph, accounts, i, visited, emailSet);
                res.resize(res.size() + 1);
                res.back().push_back(accounts[i][0]);
                for (const string& email : emailSet) {
                    res.back().push_back(email);
                }
            }
        }
        
        return res;
    }
};

Solution 3: Union Find

class UF {
    
private:
    vector<int> parent;
    vector<int> rank;
    int count;
    int N;
    
public:
    
    UF(int N) : parent(N), rank(N, 0), N(N), count(N) {
        for (int i = 0; i < N; i++) {
            parent[i] = i;
        }
    }
    
    int find(int p) {
        while (p != parent[p]) {
            parent[p] = parent[parent[p]];    // path compression by halving
            p = parent[p];
        }
        return p;
    }
    
    int getCount() const {
        return count;
    }
    
    bool connected(int p, int q) {
        return find(p) == find(q);
    }
    
    void Union(int p, int q) {
        int rootP = find(p);
        int rootQ = find(q);
        if (rootP == rootQ) return;
        
        // make root of smaller rank point to root of larger rank
        if (rank[rootP] < rank[rootQ]) parent[rootP] = rootQ;
        else if (rank[rootP] > rank[rootQ]) parent[rootQ] = rootP;
        else {
            parent[rootQ] = rootP;
            rank[rootP]++;
        }
        count--;
    }
};

class Solution {
public:
    vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
        int n = accounts.size();
        map<string, vector<int>> mymap;
        for (int i = 0; i < n; i++) {
            for (int j = 1; j < accounts[i].size(); j++) {
                mymap[accounts[i][j]].push_back(i);
            }
        }
        
        UF uf(n);
        for (const auto& it : mymap) {
            for (int i = 1; i < it.second.size(); i++) {
                uf.Union(it.second[0], it.second[i]);
            }
        }
        
        vector<vector<string>> result;
        unordered_map<int, int> root_2_result_idx;
        for (const auto& it : mymap) {
            int root = uf.find(it.second[0]);
            int resultSize = root_2_result_idx.size();
            if (root_2_result_idx.count(root) == 0) {
                root_2_result_idx[root] = resultSize;
                vector<string> temp;
                temp.push_back(accounts[root][0]);
                result.push_back(temp);
            }
            int index = root_2_result_idx[root];
            result[index].push_back(it.first);
        }

        return result;
    }
};