Earliest Time All Users Are Connected

You are given a list of users and a list of ride-share log entries. Each log entry records a timestamp and two users who shared a ride. Once two users share a ride, they are considered connected. Connectivity is transitive: if A is connected to B and B is connected to C, then A is connected to C.
Each log entry is formatted as "timestamp userA userB", where timestamp is an integer. Process the logs in increasing timestamp order and return the earliest timestamp when all users are connected. If the users never all become connected, return -1.

Function
earliestFullConnection(users: String[], logs: String[]) → int
Complete the function earliestFullConnection in the editor.
earliestFullConnection has the following parameters:
String users[]: all users in the system
String logs[]: ride-share connection logs
Returns int: the earliest timestamp when all users are connected, or -1 if this never happens

Examples
Example 1
users = ["A", "B", "C"]
logs = ["1 A B", "3 B C"]
return = 3
At timestamp 1, A and B are connected. At timestamp 3, C joins that component through B, so all users are connected.
Example 2
users = ["A", "B", "C", "D"]
logs = ["5 A B", "1 C D", "10 B C"]
return = 10
The logs are processed by timestamp: C-D at 1, A-B at 5, then B-C at 10 connects the two components.
Example 3
users = ["A", "B", "C"]
logs = ["2 A B"]
return = -1
User C never connects to the A-B component.

Constraints
Each log entry has the format "timestamp userA userB".
Logs may be provided out of timestamp order.
User names in logs are included in users.

Here's the trick. Parse each log into (timestamp, a, b), sort by timestamp, then walk them in order. Use a disjoint set over the users list. Each union that actually merges two different roots drops the component count by one. Start the count at the number of users. The moment it hits 1, return that log's timestamp. If you finish the loop and the count is still above 1, return -1. The common pitfalls: forgetting to sort because the constraints say logs may be out of order, sorting timestamps as strings so "10" lands before "5", and decrementing the count on a union of already-connected users. Also handle a single user, where the answer is arguably immediate. Use path compression and union by size or rank. If the live OA freezes your brain, StealthCoder can hand you the union-find skeleton so you only fix the edge cases.

StealthCoder is the hedge for the one pattern you didn't drill. It runs invisibly during the screen share.