It frees most time for the same one-course loss. Deadline order exposes violated prefixes; replacing the longest chosen job maximizes remaining room without reducing count more than necessary.

Keep a feasible-looking set through each deadline. If its total time is too large, one course must leave; evicting the longest produces the smallest possible total for that count.

What the function promises: Return the maximum course count for nonnegative durations and deadlines.

from heapq import heappush, heappop

def max_courses(courses):
    total, chosen = 0, []
    for duration, deadline in sorted(courses, key=lambda item: item[1]):
        total += duration
        heappush(chosen, -duration)
        if total > deadline:
            total += heappop(chosen)
    return len(chosen)

assert max_courses([(100,200),(200,1300),(1000,1250),(2000,3200)]) == 3
assert max_courses([]) == 0

Work and storage: O(n log n) time and O(n) heap space.

When to recognize it: Deadline order exposes violated prefixes; replacing the longest chosen job maximizes remaining room without reducing count more than necessary.

Check before applying it: This proof maximizes count, not reward, and assumes every course is available from time zero.

Count-preserving replacement is no longer enough; use a reward-aware DP or another justified objective-specific method.