"""
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 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.
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] <= 30
accounts[i][0] consists of English letters.
accounts[i][j] (for j > 0) is a valid email.
"""

class Solution:
    def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
        """
        Intuition

        Draw an edge between two emails if they occur in the same account. The problem comes down to finding connected components of this graph.

        Algorithm

        For each account, draw the edge from the first email to all other emails. Additionally, we'll remember a map from emails to names on the side. After finding each connected component using a depth-first search, we'll add that to our answer.
        otherwise update the list
        
        Time Complexity O(a log a) due to sorting of the component
        Space Complexity: O(a) where a is the number of emails in an account due to graph and stack
        """
        
        email_to_name = dict()
        graph = collections.defaultdict(set)
        for account in accounts:
            name = account[0]
            for email in account[1:]:
                graph[account[1]].add(email)
                graph[email].add(account[1])
                email_to_name[email] = name
        
        seen = set()
        ans = []
        for email in graph:
            if email not in seen:
                seen.add(email)
                stack = [email]
                component = []
                while stack:
                    node = stack.pop()
                    component.append(node)
                    for nei in graph[node]:
                        if nei not in seen:
                            seen.add(nei)
                            stack.append(nei)
                ans.append([email_to_name[email]] + sorted(component))
        return ans
