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Question
Solution

Implement a MyCalendar class to store your events. A new event can be added if adding the event will not cause a double booking.

Your class will have the 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 double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false and do not add the event to the calendar.

Your class will be called like this: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)

Example 1:

MyCalendar();
MyCalendar.book(10, 20); // returns true
MyCalendar.book(15, 25); // returns false
MyCalendar.book(20, 30); // returns true
Explanation: 
The first event can be booked.  The second can't because time 15 is already booked by another event.
The third event can be booked, as the first event takes every time less than 20, but not including 20.

Note:

The number of calls to MyCalendar.book per test case will be at most 1000.

In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

Approach #1: Brute Force [Accepted]
Approach #2: Balanced Tree [Accepted]

Approach #1: Brute Force [Accepted]

Intuition

When booking a new event [start, end), check if every current event conflicts with the new event. If none of them do, we can book the event.

Algorithm

We will maintain a list of interval events (not necessarily sorted). Evidently, two events [s1, e1) and [s2, e2) do not conflict if and only if one of them starts after the other one ends: either e1 <= s2 OR e2 <= s1. By De Morgan\'s laws, this means the events conflict when s1 < e2 AND s2 < e1.

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Complexity Analysis

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Time Complexity: $O(N^2)$ , where $N$ is the number of events booked. For each new event, we process every previous event to decide whether the new event can be booked. This leads to $\\sum_k^N O(k) = O(N^2)$ complexity.
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Space Complexity: $O(N)$ , the size of the calendar.
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Approach #2: Balanced Tree [Accepted]

Intuition

If we maintained our events in sorted order, we could check whether an event could be booked in $O(\\log N)$ time (where $N$ is the number of events already booked) by binary searching for where the event should be placed. We would also have to insert the event in our sorted structure.

Algorithm

We need a data structure that keeps elements sorted and supports fast insertion. In Java, a TreeMap is the perfect candidate. In Python, we can build our own binary tree structure.

For Java, we will have a TreeMap where the keys are the start of each interval, and the values are the ends of those intervals. When inserting the interval [start, end), we check if there is a conflict on each side with neighboring intervals: we would like calendar.get(prev)) <= start <= end <= next for the booking to be valid (or for prev or next to be null respectively.)

For Python, we will create a binary tree. Each node represents some interval [self.start, self.end) while self.left, self.right represents nodes that are smaller or larger than the current node.

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Complexity Analysis

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Time Complexity (Java): $O(N \\log N)$ , where $N$ is the number of events booked. For each new event, we search that the event is legal in $O(\\log N)$ time, then insert it in $O(1)$ time.
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Time Complexity (Python): $O(N^2)$ worst case, with $O(N \\log N)$ on random data. For each new event, we insert the event into our binary tree. As this tree may not be balanced, it may take a linear number of steps to add each event.
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Space Complexity: $O(N)$ , the size of the data structures used.
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Analysis written by: @awice. Solutions in Approach #2 inspired by @shawngao and @persianPanda.

729. My Calendar I

Approach #1: Brute Force [Accepted]

Approach #2: Balanced Tree [Accepted]