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 <= 10002 <= accounts[i].length <= 101 <= accounts[i][j].length <= 30accounts[i][0] consists of English letters.accounts[i][j] (for j > 0) is a valid email.In this approach, we treat each email as a node in a graph and create edges between nodes that appear in the same account. By doing so, connected components in the graph will represent emails belonging to the same person.
Once the graph is constructed, we perform a Depth-First Search (DFS) to enumerate all emails in each connected component, sort the collected emails, and then append the owner’s name to produce the final result.
The Python solution uses a dictionary to map emails to a graph structure. Each email is treated as a node, and accounts describe connections. We apply a depth-first search to explore connected components and collect emails of each component. Each collection is sorted and paired with the account owner’s name before appending to the result list.
C++
Time complexity is O(NK log NK), where N is the number of accounts, and K is the maximum number of emails in an account, due to sorting each component. Space complexity is O(NK) for storing the graph and visited nodes.
This approach uses a Union-Find data structure for efficient merging of accounts. Each email is identified with a parent, initially itself, and accounts are unified if they share an email. We then deduce separate email sets from parent-child relationships.
In the Java solution, Union-Find is used to connect and find associated emails. We first store each email with its own parent, then use the find operation to determine shared components. Finally, a mapping between connected components and their parent node helps in gathering and sorting emails accordingly.
JavaScript
Time complexity is O(NK log NK) due to sorting each email list. Space complexity is O(NK) owing to email mappings and Union-Find arrangements.
| Approach | Complexity |
|---|---|
| Graph-Based Approach | Time complexity is O(NK log NK), where N is the number of accounts, and K is the maximum number of emails in an account, due to sorting each component. Space complexity is O(NK) for storing the graph and visited nodes. |
| Union-Find Approach | Time complexity is O(NK log NK) due to sorting each email list. Space complexity is O(NK) owing to email mappings and Union-Find arrangements. |
G-50. Accounts Merge - DSU • take U forward • 125,577 views views
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