Combine Accounts Connected by Shared Addresses

### Problem Each account contains a name followed by email addresses. Accounts sharing an address belong to the same person, transitively. Matching names alone do not establish identity. A person's accounts have the same name. Return one row per merged person, beginning with the name and followed by unique addresses in sorted order. Merged rows may appear in any order. ```cpp class Solution { public: vector<vector<string>> accountsMerge(vector<vector<string>>& accounts); }; ``` ### Examples **Example 1:** ```text accounts = [["Ada","b@example.com","a@example.com"],["Ada","b@example.com","c@example.com"]] One valid output: [["Ada","a@example.com","b@example.com","c@example.com"]] Explanation: The shared b address connects both rows. ``` **Example 2:** ```text accounts = [["Bo","x@example.com"],["Bo","y@example.com"]] One valid output: [["Bo","x@example.com"],["Bo","y@example.com"]] Explanation: Equal names do not merge distinct address sets. ``` ### Constraints - There are 1 through 1000 account rows, each containing 2 through 10 strings. - Each string has length 1 through 30. - Names use English letters; later strings are valid email addresses. ### Contract considerations Preserve the input row schema, unique addresses and within-row sorting. ```hint Read the boundary condition Which graph connectivity relation is established by one common address? ``` ### What a Strong Answer Covers Preserve the input row schema, unique addresses and within-row sorting. ### Follow-up Questions Optional practice extension: How would you accept additional accounts incrementally while retaining exact email-based connectivity?