# My Calendar III

`Sweep Line`, `TreeMap`, `Interval`

**Hard**

Implement a `MyCalendarThree` class to store your events. A new event can **always** be added.

Your class will have one method, `book(int start, int end)`. Formally, this represents a booking on the half open interval`[start, end)`, the range of real numbers `x` such that `start <= x < end`.

AK-bookinghappens when **K** events have some non-empty intersection (ie., there is some time that is common to all K events.)

For each call to the method`MyCalendar.book`, return an integer `K` representing the largest integer such that there exists a`K`-booking in the calendar.

Your class will be called like this:

`MyCalendarThree cal = new MyCalendarThree();`

`MyCalendarThree.book(start, end)`

**Example 1:**

```
MyCalendarThree();
MyCalendarThree.book(10, 20); // returns 1
MyCalendarThree.book(50, 60); // returns 1
MyCalendarThree.book(10, 40); // returns 2
MyCalendarThree.book(5, 15); // returns 3
MyCalendarThree.book(5, 10); // returns 3
MyCalendarThree.book(25, 55); // returns 3

Explanation:

The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
The remaining events cause the maximum K-booking to be only a 3-booking.
Note that the last event locally causes a 2-booking, but the answer is still 3 because
eg. [10, 20), [10, 40), and [5, 15) are still triple booked.
```

**Note:**

* The number of calls to `MyCalendarThree.book` per test case will be at most `400`.
* In calls to `MyCalendarThree.book(start, end)`, `start` and `end` are integers in the range `[0, 10^9]`.

## Solution & Analysis

理解 k-booking：

> A K-booking happens when **K** events have some non-empty intersection (ie., there is some time that is common to all K events.)

### Approach: Boundary Count - TreeMap - Sweep Line

基本思想和My Calendar II一样，这里问的变成了k-booking的值，每次新加入的booking不会被拒绝，那也就是说返回所有booking的重叠次数的最大值。完全套用TreeMap based扫描线做法即可。

**Time**: each booking() operation - O(n + logn), total for n booking(): O(n^2 + nlogn) \~ O(n^2)

**Space**: O(n)

```java
class MyCalendarThree {
    TreeMap<Integer, Integer> calendar;

    public MyCalendarThree() {
        calendar = new TreeMap<>();
    }

    public int book(int start, int end) {
        calendar.put(start, calendar.getOrDefault(start, 0) + 1);
        calendar.put(end, calendar.getOrDefault(end, 0) - 1);

        int ongoing = 0;
        int maxBooking = 0;
        for (int v: calendar.values()) {
            ongoing += v;
            https://leetcode.com/problems/my-calendar-iii/ = Math.max(maxBooking, ongoing);
        }
        return maxBooking;
    }
}

/**
 * Your MyCalendarThree object will be instantiated and called as such:
 * MyCalendarThree obj = new MyCalendarThree();
 * int param_1 = obj.book(start,end);
 */
```

## Reference

[https://leetcode.com/problems/my-calendar-iii/discuss/109556/JavaC%2B%2B-Clean-Code](https://leetcode.com/problems/my-calendar-iii/discuss/109556/JavaC%2B%2B-Clean-Code)
