721. Accounts Merge (Python)

Related Topic

Depth-First-Search. Graph.

Description

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.

Sample I/O

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.

Constraints:

• 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].

Methodology

Build up a dictionary which key is email address and value is a list of the names (or index) that use this email. The iterate the given account and use dfs to find each name (or index)’s emails. The dfs will recursively append the emails of given name and find other name (or index) as given name based on neighbour emails. The key point here is same eamil belong to same person.

Code

visited_accounts = [False] * len(accounts)
        emails_accounts_map =collections.defaultdict(list)
        res = []
        # Build up the graph.
        for i, account in enumerate(accounts):
            for j in range(1, len(account)):
                email = account[j]
                emails_accounts_map[email].append(i)
        # DFS code for traversing accounts.
        def dfs(i, emails):
            if visited_accounts[i]:
                return
            visited_accounts[i] = True
            for j in range(1, len(accounts[i])):
                email = accounts[i][j]
                emails.add(email)
                for neighbor in emails_accounts_map[email]:
                    dfs(neighbor, emails)
        # Perform DFS for accounts and add to results.
        for i, account in enumerate(accounts):
            if visited_accounts[i]:
                continue
            name, emails = account[0], set()
            dfs(i, emails)
            res.append([name] + sorted(emails))
        return res

BigO

Time Complexity: O(∑Ai*logAi), where Ai is the length of accounts[i]. Without the log factor, this is the complexity to build the graph and search for each component. The log factor is for sorting each component at the end.