Earliest Time to Connect All Riders

Given timestamped ride logs where each entry links two riders, find the earliest timestamp at which every rider that appears anywhere in the logs belongs to a single connected component. Return "-1" if full connection never happens. A union-find / connected-component problem in the LeetCode 1101 family. Earliest Time to Connect All Riders Given timestamped ride logs where each entry links two riders, find the earliest timestamp at which every rider that appears anywhere in the logs belongs to a single connected component. Return "-1" if full connection never happens. A union-find / connected-component problem in the LeetCode 1101 family. SWE union-find graph medium Frequency Single report Last asked 2026-05-22 Stage phone-screen · onsite-coding Earliest Time to Connect All Riders You are given logs, where logs[i] = [timestamp, riderA, riderB] indicates that riderA and riderB shared a ride at timestamp. The logs are sorted in strictly increasing timestamp order. Two riders are connected if they shared a ride directly or can be linked transitively through other riders. Return the earliest timestamp when every rider that appears anywhere in logs belongs to a single connected component. If this never happens, return "-1". In this practice version, timestamps are represented as strings so the input can stay in a simple string[][] format. Because the logs are already sorted, you only need to return the first timestamp that completes the full connection. Follow-up note: If the stream can also contain block events such as ["1670000541", "Bob", "Dan", "blocked"], edges can disappear over time. Plain Union-Find no longer works online because deletions break monotonicity. Typical solutions use offline processing with rollback Union-Find or another dynamic-connectivity technique. Examples Example 1: Input: logs = [["1","Alice","Boo"],["2","Charlie","Dan"],["3","Bob","Charlie"],["4","Alice","Evel"],["5","Bob","Alice"]] Output: "5" Explanation: At timestamp 5, the edge between Bob and Alice merges the two remaining groups into one connected component. Example 2: Input: logs = [["1","Alice","Boo"],["2","Charlie","Dan"],["3","Bob","Charlie"],["4","Alice","Evel"],["5","Bob","Dan"]] Output: "-1" Explanation: Even after the final log, {Alice, Boo, Evel} and {Bob, Charlie, Dan} remain disconnected. Constraints 1 <= logs.length <= 10^5 logs[i].length == 3 1 <= timestamp.length, riderA.length, riderB.length <= 20 riderA != riderB logs is sorted by strictly increasing timestamp