Top-K Heap keeps the best k items and evicts the weakest when there are too many. This card has no fixed k. The limit is a deadline: the courses taken must finish by each last_day. The card evicts the longest taken course when the limit breaks.

Courses are sorted by last_day. Taken courses run one after another, so after the course with last_day the total time is total_time. If total_time is at most last_day, every taken course so far finishes in time, because earlier courses have earlier or equal last days and finish earlier.

If total_time is more than last_day, one course must go. Dropping the longest course taken so far gives the smallest total_time that is possible with the same number of courses. A smaller total_time leaves the most room for later courses. If the new course is the longest, it is the one dropped, and the set stays as it was.

The board uses the courses [4, 7], [5, 6], [2, 14], [4, 11], [5, 10], [4, 13]. Sorted by last_day, they are [5, 6], [4, 7], [5, 10], [4, 11], [4, 13], [2, 14]. The max_heap regrets two courses of 5 days, and schedule_course returns 4. Course Schedule III (LC 630) has the same shape.

The sort key must be last_day. On the board's courses, sorting by duration ends with 3 courses, but the correct answer is 4. A short course with a late deadline is taken first, and it uses time that a course with an early deadline needed.

A second wrong idea is to skip a course that does not fit: if total_time + duration <= last_day. A skipped course can never be taken back, and a long course taken earlier stays in the schedule. On the board's courses, this rule returns 3. The blueprint takes the new course first and drops the longest course. This keeps the count and lowers total_time.

max_heap stores negated durations, because heapq is a min-heap. The pop returns a negative number. Write longest = -heapq.heappop(max_heap), then total_time -= longest. Without the minus sign, the subtraction adds the days.

The answer is len(max_heap), the number of courses kept. total_time is the number of days they use, which is a different value.

Sorting costs O(N log N). Each course is pushed once and popped at most once, and each heap operation costs O(log N). So the time is O(N log N) and the space is O(N).

Furthest Building You Can Reach (LC 1642) uses the same take first, regret later idea. Each climb is first paid for with a ladder. When the ladders run out, the smallest climb in the heap is paid for with bricks instead. There the heap is a min-heap, because the regret is the smallest climb.

Minimum Number of Refueling Stops (LC 871) takes every station passed, adds its fuel to a max-heap, and uses the largest fuel only when the car would otherwise stop. The heap holds choices that are made later, when they are needed.

The regret is possible only because of an exchange argument: swapping a longer taken course for a shorter new one never makes the schedule worse. A problem where a choice cannot be swapped this way, such as Non-overlapping Intervals (LC 435), is sorted by end time and has no regret step.