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

A K-booking happens 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].

Approach #1: C++.

class MyCalendarThree {
public:
    MyCalendarThree() {
        
    }
    
    int book(int start, int end) {
        ++books[start];
        --books[end];
        int count = 0;
        int ant = 0;
        for (auto it : books) {
            count += it.second;
            ant = max(ant, count);
            if (it.first > end) break;
        }
        maxNum = max(maxNum, ant);
        return maxNum;
    }
    
private:
    map<int, int> books;
    int maxNum = 0;
};

Approach #2: C++.

class MyCalendarThree {
public:
    MyCalendarThree() {
        books[INT_MAX] = 0;
        books[INT_MIN] = 0;
        maxCount = 0;
    }
    
    int book(int start, int end) {
        auto l = prev(books.upper_bound(start));
        auto r = books.lower_bound(end);
        
        for (auto curr = l, next = curr; curr != r; curr = next) {
            ++next;
            if (next->first > end) 
                books[end] = curr->second;
            if (curr->first <= start && next->first > start) {
                maxCount = max(maxCount, books[start] = curr->second+1);
            }
            else {
                maxCount = max(maxCount, ++curr->second);
            }
        }
        return maxCount;
    }
    
private:
    map<int, int> books;
    int maxCount;
};

Note:

std::prev in C++.

Approach #3: C++. [segment tree]

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