721. Accounts Merge (Medium)


Given a list accounts, 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 email that is common 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:

accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
The first and third John's are the same person as they have the common email "johnsmith@mail.com".
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', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.


  • The length of accounts will be in the range [1, 1000].
  • The length of accounts[i] will be in the range [1, 10].
  • The length of accounts[i][j] will be in the range [1, 30].
  • Solutions

    class Solution {
        // key email, value username
        private Map<String, String> emailNameMap = new HashMap<>();
        // key email, vaule connected points
        // looks like node0<->node1, node1<->node2, instead of node0<->node1, node0<->node2
        private Map<String, Set<String>> graph = new HashMap<>();
        private Set<String> done = new HashSet<>();
        public List<List<String>> accountsMerge(List<List<String>> accounts) {
            for (List<String> account : accounts) {
                String name = account.get(0);
                // previous account id
                String prev = null;
                int size = account.size();
                for (int i = 1; i < size; i++) {
                    String email = account.get(i);
                    emailNameMap.put(email, name);
                    // current email
                    String vertex = email;
                    graph.putIfAbsent(vertex, new HashSet<>());
                    // the first node should be the starting point
                    if (i != 1) {
                        //make two-way linked list
                    prev = vertex;
            List<List<String>> ans = new ArrayList<>();
            for (String email : graph.keySet()) {
                if (done.contains(email)) {
                Set<String> list = dfs(email);
                List<String> account = new ArrayList<>();
                for (String e : list) {
                //Add account user name
                String name = emailNameMap.get(account.get(0));
                account.add(0, name);
            return ans;
        private Set<String> dfs(String email) {
            Stack<String> stack = new Stack<>();
            Set<String> ans = new HashSet<>();
            while(stack.size() > 0) {
                String x = stack.pop();
                Set<String> children = graph.get(x);
                if (children == null) {
                for (String c : children) {
                    if (!done.contains(c)) {
            return ans;

    Incorrect Solutions


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