731. My Calendar Ii¶

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731. My Calendar II

Medium

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.

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.

Constraints:

• 0 <= start < end <= 109
• At most 1000 calls will be made to book.

Solution¶

class MyCalendarTwo {
    private TreeMap<Integer, Integer> bookings;
    public MyCalendarTwo() {
        bookings = new TreeMap<>();
    }
    public boolean book(int start, int end) {
        bookings.put(start, bookings.getOrDefault(start, 0) + 1);
        bookings.put(end, bookings.getOrDefault(end, 0) - 1);
        int cnt = 0;
        for (int count : bookings.values()) {
            cnt += count;
            if (cnt >= 3) {
                bookings.put(start, bookings.get(start) - 1);
                if (bookings.get(start) == 0) bookings.remove(start);
                bookings.put(end, bookings.get(end) + 1);
                if (bookings.get(end) == 0) bookings.remove(end);
                return false;
            }
        }
        return true;
    }
}

/**
 * Your MyCalendarTwo object will be instantiated and called as such:
 * MyCalendarTwo obj = new MyCalendarTwo();
 * boolean param_1 = obj.book(start,end);
 */

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here