Exercise 3 solution

    from datetime import datetime, timedelta
    from collections import Counter

    EPOCH = datetime(2000, 1, 1)

    def floor_to(dt, step):
        return EPOCH + ((dt - EPOCH) // step) * step

    def hms(td):
        s = int(td.total_seconds())
        sign = "-" if s < 0 else ""
        h, r = divmod(abs(s), 3600)
        m, s = divmod(r, 60)
        return f"{sign}{h}:{m:02d}:{s:02d}"

    events = [datetime(2026, 11, 2, 9, 5), datetime(2026, 11, 2, 9, 14, 59),
              datetime(2026, 11, 2, 9, 15), datetime(2026, 11, 2, 9, 44),
              datetime(2026, 11, 2, 10, 0), datetime(2026, 11, 2, 10, 1),
              datetime(2026, 11, 2, 23, 59, 59)]
    counts = Counter(floor_to(e, timedelta(minutes=15)) for e in events)
    for k in sorted(counts):
        print(k.strftime("%H:%M"), counts[k])
    span = max(events) - min(events)
    print(span, hms(span), hms(-span), hms(timedelta(seconds=-1)))
    print(floor_to(datetime(2026, 11, 2, 9, 14, 59), timedelta(days=1)))

Output (run, and exactly as shown):

    09:00 2
    09:15 1
    09:30 1
    10:00 2
    23:45 1
    14:54:59 14:54:59 -14:54:59 -0:00:01
    2026-11-02 00:00:00

Explanation: floor_to measures from a fixed anchor, so 09:14:59 stays in the
09:00 bucket and 09:15:00 starts the next. The bucket at 09:30 holds 09:44 and
10:00 starts a new hour, so 10:00 and 10:01 share a bucket. The span is the
latest minus the earliest, 14:54:59, and hms keeps counting hours. Negating it
puts the sign at the front, and minus one second prints as -0:00:01 (str() would
have shown -1 day, 23:59:59). Flooring to a whole day gives midnight.