A k
-booking happens when k
events have some non-empty intersection (i.e., there is some time that is common to all k
events.)
You are given some events [start, end)
, after each given event, return an integer k
representing the maximum k
-booking between all the previous events.
Implement the MyCalendarThree
class:
MyCalendarThree()
Initializes the object.int book(int start, int end)
Returns an integerk
representing the largest integer such that there exists ak
-booking in the calendar.
Example 1:
Input ["MyCalendarThree", "book", "book", "book", "book", "book", "book"] [[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]] Output [null, 1, 1, 2, 3, 3, 3] Explanation MyCalendarThree myCalendarThree = new MyCalendarThree(); myCalendarThree.book(10, 20); // return 1, The first event can be booked and is disjoint, so the maximum k-booking is a 1-booking. myCalendarThree.book(50, 60); // return 1, The second event can be booked and is disjoint, so the maximum k-booking is a 1-booking. myCalendarThree.book(10, 40); // return 2, The third event [10, 40) intersects the first event, and the maximum k-booking is a 2-booking. myCalendarThree.book(5, 15); // return 3, The remaining events cause the maximum K-booking to be only a 3-booking. myCalendarThree.book(5, 10); // return 3 myCalendarThree.book(25, 55); // return 3
Constraints:
0 <= start < end <= 109
- At most
400
calls will be made tobook
.