731. My Calendar II

1. 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 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.

2. Example

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 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.

3. Constraints

• 0 <= start < end <= $10^9$
• At most 1000 calls will be made to book.

4. Solutions

Boundary Count

n is the times of booking
Time complexity: O($n^2$)
Space complexity: O(n)

class MyCalendarTwo {
public:
    MyCalendarTwo() {}

    bool book(int start, int end) {
        ++count_diff_[start];
        --count_diff_[end];
        int max_book = 0;
        for (auto &[node, count] : count_diff_) {
            max_book += count;
            if (node >= end) {
                break;
            }
            if (max_book > order_limit_) {
                --count_diff_[start];
                ++count_diff_[end];

                return false;
            }
        }

        return true;
    }

private:
    const int order_limit_ = 2;
    map<int, int> count_diff_;
};