#731

My Calendar II

medium · 62.9% accepted · 2,270 likes · top 65%

array · binary search · design · segment tree · prefix sum · ordered set

Description

You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a triple booking.

A triple booking happens when three events have some non-empty intersection (i.e., some moment is common to all the three events.).

The event can be represented as a pair of integers startTime and endTime that represents a booking on the half-open interval [startTime, endTime), the range of real numbers x such that startTime <= x < endTime.

Implement the MyCalendarTwo class:

- MyCalendarTwo() Initializes the calendar object.

- boolean book(int startTime, int endTime) Returns true if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return false and do not add the event to the calendar.

Example 1:

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]

Example 2:

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.

Solution