First Timestamp When All People Are Connected

### Problem People are labeled 0 through n-1. A log `[timestamp,x,y]` adds a symmetric friendship between x and y, which persists. Acquaintance includes friendship chains. Return the earliest logged timestamp when every person is connected to every other, or -1 if this never happens. ```cpp class Solution { public: int earliestAcq(vector<vector<int>>& logs, int n); }; ``` ### Examples **Example 1:** ```text Input: logs = [[8,0,1],[3,1,2]], n = 3 Output: 8 Explanation: At3 only1 and2 are connected; the event at8 connects person0. ``` **Example 2:** ```text Input: logs = [[1,0,1],[2,2,3]], n = 4 Output: -1 Explanation: The two pairs remain in separate components. ``` ### Constraints - `2 <= n <= 100`; the log list has 1 to 10000 entries. - Every log contains exactly three integers. - Timestamps lie in `[0,1000000000]` and are all distinct. - Each endpoint lies in `[0,n-1]`, with distinct endpoints in a log. - Each endpoint pair occurs at most once. ### Contract considerations Timestamps are numeric event labels, not a required calendar format. ```hint Read the boundary condition Could a later-listed log have an earlier timestamp, and how does that affect the meaning of the requested earliest time? ``` ### Boundary reminders Timestamps are numeric event labels, not a required calendar format.