The two-heap setup clicked for me pretty quickly but I fumbled the ordering on the busy heap at first.
Model the problem as a simulation with two priority queues: one min-heap for available workers keyed by worker index, and one min-heap for busy workers keyed by their free time. Process tasks in order of arrival, first releasing any workers whose free time is <= the task's arrival time, then assign the task to the smallest-index available worker or enqueue it if none are free.
Pro tip: Clarify edge cases upfront—like simultaneous task arrivals, tasks arriving before any worker is free, or multiple workers freeing at the same time—and state your assumptions. This shows thoroughness and prevents miscommunication.
Ask about input format, tie-breaking rules (e.g., if multiple workers free at the same time, pick lowest index), and whether tasks are processed in arrival order. Confirm output format.
Use a min-heap for available workers (by index) and a min-heap for busy workers (by free time). This allows O(log N) operations for releasing and assigning workers.
Iterate through tasks sorted by arrival time. For each task, release all workers whose free time <= task arrival time into the available heap. If available heap is non-empty, pop the smallest index and assign; otherwise, add the task to a queue.
After processing all arrivals, process any queued tasks in FIFO order. When a worker becomes free, assign the next queued task to the lowest-index available worker, updating the worker's free time.
State time complexity O((N+M) log N) and space O(N+M). Walk through a small example to verify correctness, including edge cases like no available workers initially.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.