## # 732. My Calendar III

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.
``````

### # Solution

Approach 1: use a map to keep all times that events(start or end) happen.

Approach 2: keep intervals, and go through them while keeping a heap for active/in flight intervals.

### # Code (Python)

Approach 1:

``````class MyCalendarThree:

def __init__(self):
self.events = {}

def book(self, start: int, end: int) -> int:
# O(NlogN) time
if start not in self.events:
self.events[start] = 0
self.events[start] += 1
if end not in self.events:
self.events[end] = 0
self.events[end] -= 1

most_books = 0
current_books = 0
for time in sorted(self.events.keys()):
current_books += self.events[time]
most_books = max(most_books, current_books)

return most_books
``````

Approach 2:

``````class MyCalendarThree:
def __init__(self):
self.intervals = []

def book(self, start: int, end: int) -> int:
# O(NlogN) in the worst case
index = bisect.bisect(self.intervals, [start, end])
if index < len(self.intervals):
self.intervals = self.intervals[:index] + [[start, end]] + self.intervals[index:]
else:
self.intervals = self.intervals + [[start, end]]

# heap keeps the end times of all intervals in flight
heap = []
heapq.heapify(heap)
most_books = 0
for interval in self.intervals:
while heap and heap[0] <= interval[0]:
heapq.heappop(heap)
heapq.heappush(heap, interval[1])
most_books = max(most_books, len(heap))
return most_books
``````

Approach 1:

Approach 2: