I recognized this as basically LeetCode 855 but without the leave() part, which actually made it simpler.
Clarify the problem constraints and edge cases, then propose an efficient solution using a priority queue to track available intervals, prioritizing by maximum minimum distance. Walk through the algorithm step-by-step, analyze time and space complexity, and discuss potential optimizations or alternative approaches.
Pro tip: Mention that this is essentially the 'Exam Room' problem (LeetCode 855) and that using a priority queue with a custom comparator yields O(log n) per seat assignment, which is optimal for this problem.
Ask clarifying questions about n (range, type), whether seats are assigned one at a time, and if there are any constraints on time complexity. Confirm that the first seat is always 0 and seats are never vacated.
Use a priority queue (max-heap) to store available intervals, prioritized by the minimum distance to occupied seats. Also maintain a set or boolean array to track occupied seats for quick lookup.
For an interval [left, right] of unoccupied seats, the best seat is the one that maximizes the minimum distance to occupied seats. Compute the distance as min(seat - left, right - seat) and prioritize intervals by this distance, breaking ties by smaller seat index.
When assigning a seat, pop the interval with the highest priority, choose the seat that maximizes the minimum distance (usually the middle), mark it occupied, and push the resulting left and right sub-intervals back into the priority queue.
Each seat assignment takes O(log n) time due to heap operations, and space is O(n). Test with small n, n=1, and sequences of assignments to ensure correctness.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.