721 Accounts Merge

721. Accounts Merge

1. Question

Given a listaccounts, each elementaccounts[i]is a list of strings, where the first elementaccounts[i][0]is aname, and the rest of the elements areemailsrepresenting 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:

Input:

accounts = [["John", "[e-mail]", "[e-mail]"], ["John", "[e-mail]"], ["John", "[e-mail]", "[e-mail]"], ["Mary", "[e-mail]"]]

Output:
 [["John", '[e-mail]', '[e-mail]', '[e-mail]'],  ["John", "[e-mail]"], ["Mary", "[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.

Note:

The length ofaccountswill be in the range[1, 1000].

The length ofaccounts[i]will be in the range[1, 10].

The length ofaccounts[i][j]will be in the range[1, 30].

2. Implementation

(1) Union Find

class Solution {
    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        List<List<String>> res = new ArrayList();

        if (accounts == null || accounts.size() == 0) {
            return res;
        }

        UnionFind uf = new UnionFind(accounts.size());

        Map<String, Integer> emailToUser = new HashMap();

        for (int user = 0; user < accounts.size(); user++) {
            List<String> account = accounts.get(user);

            for (int i = 1; i < account.size(); i++) {
                String email = account.get(i);

                emailToUser.putIfAbsent(email, user);
                uf.union(emailToUser.get(email), emailToUser.get(account.get(1)));
            }
        }

        Map<Integer, List<String>> userToEmails = new HashMap();

        for (String email : emailToUser.keySet()) {
            int user = uf.find(emailToUser.get(email));
            userToEmails.putIfAbsent(user, new ArrayList());
            userToEmails.get(user).add(email);
        }

        for (int user : userToEmails.keySet()) {
            List<String> emails = userToEmails.get(user);
            Collections.sort(emails);
            emails.add(0, accounts.get(user).get(0));
            res.add(emails);
        }
        return res;
    }

    class UnionFind {
        int[] parent;
        int[] size;
        int count;

        public UnionFind(int n) {
            parent = new int[n];
            size = new int[n];
            count = n;

            for (int i = 0; i < n; i++) {
                parent[i] = i;
                size[i] = 1;
            }
        }

        public int find(int node) {
            while (node != parent[node]) {
                node = parent[node];
            }
            return node;
        }

        public void union(int i, int j) {
            int root1 = find(i);
            int root2 = find(j);

            if (root1 == root2) return;

            if (size[root1] < size[root2]) {
                parent[root1] = root2;
                size[root2] += size[root1];
            }
            else {
                parent[root2] = root1;
                size[root1] += size[root2];
            }
            --count;
        }
    }
}

3. Time & Space Complexity

(1) Union Find: 时间复杂度O(mnlogn), m是user个数, n是每个user的email个数，最主要的时间是从userToEmails中, 对每个user的email进行排序，空间复杂度O(mn)