Accounts Merge

Problem:

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:

Input: 
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"]]
Explanation: 
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.

Note:

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].

Analysis：

Union find

1. map email string to integer.
2. connect email to its account holder's name
3. construct the union find structure
4. group emails, each father of the group as key
5. construct result from the email group

Solution:

这道题感觉挺难，如何把链接好的email，一组一组取出来是个挑战。Hashtable 如果value是一个collection可以考虑computeIfAbsent.

class Solution {
    class UnionFind {
        int[] father;
        public UnionFind() {
            father = new int[10000];
            for (int i = 0; i < 10000; i++) {
                father[i] = i;
            }
        }

        public int find(int n) {
            if (father[n] != n) father[n] = find(father[n]);
            return father[n];
        }

        public void union(int a, int b) {
            father[find(a)] = find(b);
        }

        public boolean query(int a, int b){
            return find(a) == find(b);
        }
    } 

    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        if (accounts == null || accounts.size() == 0)   
            return null;
        Map<String, String> emailToName = new HashMap<>();
        Map<String, Integer> emailToId = new HashMap<>();
        UnionFind uf =  new UnionFind();
        int count = 0;
        for (List<String> acc: accounts) {
            for (int i = 1; i < acc.size(); i++) {
                // map to int
                if (acc.get(i) == "")
                    continue;
                if (!emailToId.containsKey(acc.get(i))) {
                    emailToId.put(acc.get(i), count++);
                }
                
                emailToName.put(acc.get(i), acc.get(0));
                
                // self union
                uf.union(emailToId.get(acc.get(1)), emailToId.get(acc.get(i)));
            }
        }

        Map<Integer, List<String>> groups = new HashMap<>();
        // group emails
        for (Map.Entry<String, Integer> entry: emailToId.entrySet() ){
            int father = uf.find(entry.getValue());
            groups.computeIfAbsent(father, k -> new ArrayList<String>()).add(entry.getKey());
        }
        List<List<String>> res = new ArrayList<>();
        for (Map.Entry<Integer, List<String>> entry: groups.entrySet()) {
            Collections.sort(entry.getValue());
            List<String> temp = new ArrayList<String>(entry.getValue());
            temp.add(0, emailToName.get(entry.getValue().get(0)));
            res.add(temp);
        }
        return res;
    }
}