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Looking at the boarding passes (A, B) (B, C)(C, D)(D, F), we can see only the departure and destination cities appear once. And the other cities appear twice.
- roysongbb10 October 17, 2015So, we can build a HashMap to store the city name as key and departure/destination(boolean) as value.
Iterate all boarding passes, for each city, if it isn't contained in the hashmap, store it, otherwise, remove it.
Only the departure and destination cities leave in the hashmap after visiting all boarding passes. We are done.
It is O(n) time.