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Java-based LeetCode algorithm problem solutions, regularly updated
731\. My Calendar II
Medium
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 `start` and `end` that represents a booking on the half-open interval `[start, end)`, the range of real numbers `x` such that `start <= x < end`.
Implement the `MyCalendarTwo` class:
* `MyCalendarTwo()` Initializes the calendar object.
* `boolean book(int start, int end)` 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]
**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 ca not 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.
**Constraints:**
* 0 <= start < end <= 109
* At most `1000` calls will be made to `book`.