Assign support tickets to agents by priority and deadline.

How to answer

The heap is easy. The grade is ordering, tie-breaks, and tickets that change while waiting.

1. Pin the events. add_ticket(t), agent_ready(a), close(ticket_id), and assign() -> list[(agent, ticket)] after each. Ask whether agents hold one ticket or several up to a cap.
2. Say the ordering rule out loud. “Priority first, then earliest deadline, then arrival order”: arrival is the fair tie-break. Ask: does a low-priority ticket near breach outrank a high-priority one with days left? Strict priority and earliest deadline fix the key at push time: one heap. Promotion near a breach is not: its key changes with the clock, and a heap never re-sorts. Use two heaps, by priority and by deadline: take the deadline top if inside the window, else the priority top.
3. Say what fair means for agents. Among agents under their cap: fewest open tickets, then idle longest, then lowest id, so runs repeat. Load changes on every close, so skip stale entries here too.
4. Build keys so ties never compare tickets. Push (priority_rank, deadline, seq, ticket_id), with seq from a counter and rank 0 most urgent, because heapq pops the smallest. A missing deadline is datetime.max, never None, which raises on the first tie. If your deadlines carry a time zone, use datetime.max.replace(tzinfo=timezone.utc), because a naive and an aware datetime raise on comparison too. Without seq, ties compare objects and raise, or compare ids and lose arrival order. The heapq notes cover the counter and step 5’s removed-entry marks.
5. Handle changes with lazy deletion. Closed, assigned and reprioritized tickets stay in the heaps; map each id to its current entry, skip stale ones on pop, and rebuild once they outnumber live ones.
6. Test the rule, not the heap. Equal tickets leave in arrival order, a reprioritized one jumps, a closed one never leaves, all under a fixed clock.

The trap is strict priority with no word about starvation: name it and offer aging.