accountsMerge.java

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.


INTUITION,build a graph and find connected componenets!!
Note, can also use Union find, which is an algorithm for finding connected copoents 

//TC: O(ailogai) where ai is the length of accounts of a[i]. we have to sort each componenet at end  
//SC: O(ai) where ai is the space used by our graph and search

class Solution {
    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        Map<String, Set<String>> graph = new HashMap<>();  //<email node, neighbor nodes>
        Map<String, String> name = new HashMap<>();        //<email, username>
        
        // Build the graph;
        for (List<String> account : accounts) {
            String userName = account.get(0);
            for (int i = 1; i < account.size(); i++) {
                if (!graph.containsKey(account.get(i))) {
                    graph.put(account.get(i), new HashSet<>());
                }
                name.put(account.get(i), userName);
                
                if (i == 1) continue;
                graph.get(account.get(i)).add(account.get(i - 1)); //connect the neighbors
                graph.get(account.get(i - 1)).add(account.get(i)); //connect the nodes 
            }
        }
       
        Set<String> visited = new HashSet<>();
        List<List<String>> res = new LinkedList<>();
        // DFS search the graph;
        for (String email : name.keySet()) {
            List<String> list = new LinkedList<>();
            if (visited.add(email)) { // we havent visited this email 
                dfs(graph, email, visited, list); //populate the list 
                Collections.sort(list); // order the list 
                list.add(0, name.get(email));
                res.add(list);
            }
        }
        return res;
    }
    
    public void dfs(Map<String, Set<String>> graph, String email, Set<String> visited, List<String> list) {
        list.add(email);
        for (String next : graph.get(email)) { //neighobrs 
            if (visited.add(next)) {
                dfs(graph, next, visited, list);
            }
        }
    }
}