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Accounts Merge

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 Explanation: Accounts Merge

This problem involves merging email accounts that belong to the same person. Two accounts belong to the same person if they share at least one email address. The solution uses a Union-Find data structure combined with a hash table for efficient merging and retrieval.

Approach:

  1. Union-Find: This data structure is perfect for representing and merging groups (in this case, accounts belonging to the same person). Each account is initially its own group. When we find two accounts sharing an email, we use the Union-Find to merge their groups.

  2. Hash Table (Dictionary): A hash table (d in the code) stores email addresses as keys and the index of the account containing that email as values. This allows for quick lookups to check if an email already exists and to perform unions.

  3. Merging: We iterate through the accounts. For each account and its emails:

    • If an email is already in the hash table, we merge the current account's group (using Union-Find) with the group of the account where the email was previously seen.
    • If the email is new, we add it to the hash table with the current account's index.

  4. Grouping: After merging, we need to collect all emails for each person. Another hash table (g in the code) stores the root index of each person's group (obtained from the Union-Find find() function) as a key, and a set of emails belonging to that person as the value.

  5. Result: Finally, we construct the result. For each group, we retrieve the name from the original accounts list using the root index and append the sorted set of emails.

Time Complexity:

  • The initial iteration through all accounts and emails takes O(N*M), where N is the number of accounts, and M is the average number of emails per account.
  • Union-Find operations (find() and union()) take amortized O(α(N)) time, where α(N) is the inverse Ackermann function, which grows extremely slowly and can be considered practically constant. Therefore, the union-find operations contribute negligible time.
  • Sorting the emails for each group takes O(M log M) on average. Since we do this for each group, the overall time complexity is dominated by O(NM + NM log M), which can be simplified to O(N*M log M) in the worst case (when all accounts are merged).

Space Complexity:

  • The Union-Find data structure uses O(N) space.
  • The hash tables (d and g) use O(N*M) space in the worst case (when all emails are unique).
  • The final result list can also take O(N*M) space.
  • Therefore, the overall space complexity is O(N*M).

Code Examples:

The code examples provided in Python, Java, C++, Go, and TypeScript implement this approach. Each example showcases the Union-Find implementation, hash table usage, and the merging and result-building steps. The key data structures and algorithms are consistent across all languages. Note that minor variations exist due to the specifics of each language's standard library.