Account Merge Graph DSU Solution - JAVA

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

Example 2:

Input: accounts = [["Gabe","Gabe0@m.co","Gabe3@m.co","Gabe1@m.co"],["Kevin","Kevin3@m.co","Kevin5@m.co","Kevin0@m.co"],["Ethan","Ethan5@m.co","Ethan4@m.co","Ethan0@m.co"],["Hanzo","Hanzo3@m.co","Hanzo1@m.co","Hanzo0@m.co"],["Fern","Fern5@m.co","Fern1@m.co","Fern0@m.co"]]
Output: [["Ethan","Ethan0@m.co","Ethan4@m.co","Ethan5@m.co"],["Gabe","Gabe0@m.co","Gabe1@m.co","Gabe3@m.co"],["Hanzo","Hanzo0@m.co","Hanzo1@m.co","Hanzo3@m.co"],["Kevin","Kevin0@m.co","Kevin3@m.co","Kevin5@m.co"],["Fern","Fern0@m.co","Fern1@m.co","Fern5@m.co"]]

Constraints:

• 1 <= accounts.length <= 1000

• 2 <= accounts[i].length <= 10

• 1 <= accounts[i][j].length <= 30

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

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

Solution

class Solution {
    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        class DSU{
            int[] parent;
            int[] rank;
            DSU(int n){
                this.parent = new int[n];
                this.rank = new int[n];

                for(int i=0; i<n; i++){
                    parent[i] = i;
                    rank[i] = 0;
                }
            }
            void union(int x, int y){
                int x_parent = find(x);
                int y_parent = find(y);
                if(rank[x_parent]>rank[y_parent]){
                    parent[y_parent] = x_parent;
                }else if(rank[y_parent]> rank[x_parent]){
                    parent[x_parent] = y_parent;
                }else{
                    parent[y_parent] = x_parent;
                    rank[x_parent] = rank[x_parent] + 1;
                }
            }

            int find(int x){
                if(x == parent[x]){
                    return x;
                }
                parent[x] =  find(parent[x]);
                return parent[x];
            }
        }

        int n = accounts.size();
        DSU dsu = new DSU(n);
        HashMap<String, Integer> emailMap = new HashMap<>();
        for(int i=0; i<n; i++){
            List<String> emails = accounts.get(i);
            for(int j=1; j<emails.size(); j++){
                String email = emails.get(j);
                if(!emailMap.containsKey(email)){
                    emailMap.put(email, i);   // put the parents index
                }else{
                    dsu.union(i, emailMap.get(email)); // union the current parent & the parent it has in map
                }
            }
        }

        // List parent ---> emails
        HashMap<Integer, List<String>> merged  = new HashMap<>();
        for(String email:emailMap.keySet()){
            int parent = dsu.find(emailMap.get(email)); // find the parent index
            merged.putIfAbsent(parent, new ArrayList<>()); // if parent key is not in map then create one and attach empty list
            merged.get(parent).add(email); //add the correspoding email according to the parent
        }

        //create answer list
        List<List<String>> ans = new ArrayList<>();
        for(int parent: merged.keySet()){
            List<String> emails = merged.get(parent);
            Collections.sort(emails);
            List<String> account = new ArrayList<>();
            account.add(accounts.get(parent).get(0)); //get the first element of the given accounts array which is the name
            account.addAll(emails); //after name add all the names
            ans.add(account);
        }
        return ans;
    }
}

Time & Space Complexity

Interview Summary

Operation	Time
Build email map + DSU	O(Mα(N))
Group emails	O(Mα(N))
Sort emails	O(Mlog⁡M)
Build answer	O(M)
Total	O(Mlog⁡M)

Data Structure	Space
DSU arrays	O(N)
Email map	O(M)
Grouped emails	O(M)
Answer	O(M)
Total	O(N+M)