My Calendar II Python Solution with Tests

lcpy gen -n 731   # by problem number
lcpy gen -s my_calendar_ii   # by problem name

Input
["MyCalendarTwo", "book", "book", "book", "book", "book", "book"]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
Output
[null, true, true, true, false, true, true]

Explanation
MyCalendarTwo myCalendarTwo = new MyCalendarTwo();
myCalendarTwo.book(10, 20); // return True, The event can be booked.
myCalendarTwo.book(50, 60); // return True, The event can be booked.
myCalendarTwo.book(10, 40); // return True, The event can be double booked.
myCalendarTwo.book(5, 15);  // return False, The event cannot be booked, because it would result in a triple booking.
myCalendarTwo.book(5, 10);  // return True, The event can be booked, as it does not use time 10 which is already double booked.
myCalendarTwo.book(25, 55); // return True, The event can be booked, as the time in [25, 40) will be double booked with the third event;
                            // the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

class MyCalendarTwo:
    # Tracks single and double bookings; a new event is valid iff it
    # overlaps no double-booked region.
    # Time: O(n) per book
    # Space: O(n)
    def __init__(self) -> None:
        self._booked: list[tuple[int, int]] = []
        self._double_booked: list[tuple[int, int]] = []

    @staticmethod
    def _intersects(s1: int, e1: int, s2: int, e2: int) -> bool:
        return max(s1, s2) < min(e1, e2)

    def book(self, start_time: int, end_time: int) -> bool:
        if any(self._intersects(start_time, end_time, s, e) for s, e in self._double_booked):
            return False
        for s, e in self._booked:
            if self._intersects(start_time, end_time, s, e):
                self._double_booked.append((max(start_time, s), min(end_time, e)))
        self._booked.append((start_time, end_time))
        return True