Start by clarifying requirements (e.g., single vs. multiple rooms, concurrency, time granularity) and then propose a data structure that efficiently finds an available room. For each room, maintain a sorted list of bookings (e.g., using a balanced BST or sorted array) to check for overlaps in O(log n) time. If no room is available, return a sentinel value like -1 or null.
Pro tip: Discuss how to handle concurrent booking requests to avoid double-booking, such as using locks or optimistic concurrency control, and mention that in a real system you'd likely use a database with transactions.
Ask about constraints: number of rooms, expected booking frequency, time granularity, concurrency needs, and whether bookings can be modified or cancelled.
For each room, store bookings in a sorted structure (e.g., balanced BST or sorted list) to enable efficient overlap checks. Alternatively, use an interval tree per room.
Iterate through rooms, and for each, check if the new interval overlaps with any existing booking using binary search. If no overlap, insert the booking and return the room ID.
Consider zero-length meetings, back-to-back bookings (end == start), and no available rooms. Return a sentinel value (e.g., -1) when no room is free.
Mention how to scale with many rooms/bookings (e.g., sharding by room) and how to handle concurrent requests (e.g., locking, transactions, or optimistic concurrency).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
cancelMeeting was straightforward, just a map remove.
First, clarify the data structures needed: a map from roomId to a map of start times to booking details for O(1) cancellation, and a global list or deque to track bookings in creation order for lastNScheduled. Then, implement cancelMeeting by removing the entry from the room's map and marking the booking as cancelled in the global list (or removing it if using a doubly-linked list), and implement lastNScheduled by iterating backwards from the most recent bookings, skipping cancelled ones, until N valid bookings are collected.
Pro tip: Discuss the trade-offs between eager and lazy deletion: lazy deletion (marking as cancelled) makes cancellation O(1) but requires filtering in lastNScheduled, while eager deletion (removing from a list) keeps lastNScheduled simple but may make cancellation O(n) unless using a doubly-linked list with node references. Choose based on expected read/write patterns.
Ask about expected frequency of cancellations vs. lastNScheduled calls, whether bookings can be cancelled multiple times, and if N can exceed the number of active bookings. This determines the optimal data structure.
Propose a hash map (roomId -> map of start -> booking) for O(1) cancellation, and a global doubly-linked list or dynamic array of bookings in creation order. Each booking node should have a reference to its room and start for easy removal.
Remove the booking from the room's map and from the global list. If using a doubly-linked list, unlink the node in O(1); if using an array, mark as cancelled or swap-remove with index tracking.
Traverse the global list from the end (most recent) backwards, collecting up to N non-cancelled bookings. If using lazy deletion, skip cancelled entries; if eager, just take the last N.
State time complexities: cancelMeeting O(1) with linked list, lastNScheduled O(N) or O(N + cancellations) with lazy deletion. Discuss edge cases: cancelling non-existent booking, N larger than active bookings, and concurrent modifications.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Saw this coming but still spent too long on it.
Start by clarifying the requirements and constraints (e.g., consistency vs. availability, scale, latency). Then propose a layered solution: database-level locking (optimistic or pessimistic), distributed locking for multi-service architectures, and idempotency to handle retries. Finally, discuss trade-offs and failure scenarios to show depth.
Pro tip: Mention that you'd first try to avoid distributed locks by using a single database with ACID transactions and unique constraints, as this is simpler and more reliable. Only escalate to distributed locks when necessary, and always include a fallback for lock acquisition failures.
Ask about consistency needs, expected scale, latency requirements, and whether the system is single-node or distributed. This ensures your solution fits the context.
Suggest using transactions with optimistic locking (version column) or pessimistic locking (SELECT FOR UPDATE) to prevent double booking. Highlight that this works well for single-database setups.
For microservices or multi-database scenarios, propose a distributed lock (e.g., Redis Redlock, ZooKeeper, etcd) with a lease and fencing token to ensure only one client proceeds.
Ensure booking requests are idempotent using a unique request ID, so retries don't cause double bookings. Discuss how to handle lock timeouts and client retries gracefully.
Compare approaches: optimistic vs. pessimistic locking, database vs. distributed locks, and their impact on latency, throughput, and complexity. Mention how to handle lock failures (e.g., return a conflict error).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify that 'touching endpoints are allowed' means intervals like [1,2] and [2,3] do not conflict. Then sort the intervals by start time and check if any interval's start is strictly less than the previous interval's end. If so, return false; otherwise, return true.
Pro tip: Mention that this is equivalent to detecting overlapping intervals and that sorting is the key to an O(n log n) solution. Also, discuss edge cases like empty input, single interval, and intervals with zero duration.
Confirm that intervals are closed and touching endpoints are allowed. Ask about input format and constraints.
Sort intervals by start time. Then iterate through them, comparing each start with the previous end.
If current start < previous end, return false. Otherwise, update previous end to max(previous end, current end) and continue.
Time complexity is O(n log n) due to sorting, space is O(1) if sorting in place or O(n) if creating a new list.
Walk through edge cases: empty list, single interval, touching intervals, overlapping intervals, and intervals with same start/end.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Clarify the problem constraints (e.g., intervals may be unsorted, inclusive/exclusive endpoints) and then present the sweep line algorithm: separate start and end times, sort them, and use a two-pointer technique to count concurrent meetings. Alternatively, use a min-heap to track end times while iterating through sorted intervals. Analyze time and space complexity, and discuss edge cases.
Pro tip: Mention that the minimum number of rooms equals the maximum number of overlapping intervals at any point, and that this can be computed in O(n log n) time. Also, proactively discuss how to handle edge cases like zero-length intervals or back-to-back meetings.
Ask about input format, interval inclusivity, whether intervals are sorted, and if zero-length intervals are allowed. Confirm that the goal is to find the minimum number of rooms to avoid any overlap.
Decide between the sweep line (two sorted arrays) or min-heap method. Explain that both are O(n log n) time and O(n) space, but the sweep line is simpler to implement.
For sweep line: extract start and end times, sort both arrays, then use two pointers to count active meetings, updating the maximum. For min-heap: sort intervals by start time, push end times into a min-heap, and pop when a meeting ends before the next starts.
State time complexity O(n log n) due to sorting, space O(n). Discuss edge cases: empty input, single interval, all overlapping, back-to-back meetings (end == start), and unsorted input.
Walk through a small example like [[0,30],[5,10],[15,20]] to show the algorithm yields 2 rooms. Optionally, mention that the problem is equivalent to finding the maximum number of overlapping intervals.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.