My first instinct was BFS per timestamp which would have been a mess.
Model the problem as dynamic connectivity: process events in chronological order, union the two riders in each event, and track the number of connected components. The earliest timestamp when the component count drops to 1 is the answer; if it never does, return -1.
Pro tip: Clarify assumptions upfront: whether rider IDs are known in advance, if events are sorted, and whether the graph is guaranteed to eventually connect. This shows you think about edge cases and data constraints before coding.
Ask about input format, rider ID range, event ordering, and what 'all riders' means (e.g., all riders seen in events). Confirm that events are chronological and that we need the earliest timestamp.
Use Union-Find (Disjoint Set Union) with path compression and union by rank for near-constant time operations. Maintain a count of connected components, initialized to the number of unique riders.
Iterate through events in order. For each event, union the two riders; if they were in different components, decrement the component count. After each union, check if the count equals 1.
If the component count becomes 1 at some event, return that event's timestamp. If all events are processed and the count is still >1, return -1.
Discuss time complexity O(E α(N)) and space O(N). Handle edge cases: no events, single rider, disconnected riders, duplicate events, and riders appearing only once.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.