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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 integer k
representing the largest integer such that there exists a k
-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 to book
.
[Segment Tree]
[Ordered Map]
Similar Questions
- My Calendar I (Medium)
- My Calendar II (Medium)
Hints
Hint 1
Treat each interval [start, end) as two events "start" and "end", and process them in sorted order.