731. My Calendar II

https://leetcode.com/problems/my-calendar-ii/

My Calendar II - LeetCodeLeetCode

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 ca not 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

Using two Arraylist to store the intervals, one for the all meetings, one for the overlap intervals.

If a new booking request is overlapped with one of the overlap intervals, then return false.

class MyCalendarTwo {

    ArrayList<int[]> calendar;
    ArrayList<int[]> overlaps;
    
    public MyCalendarTwo() {
        calendar = new ArrayList<>();
        overlaps = new ArrayList<>();
    }
    
    public boolean book(int start, int end) {
        for(int[] overlap : overlaps){
            if(start < overlap[1] && end > overlap[0]){
                return false;
            }
        }
        for(int[] meeting : calendar){
            if(start < meeting[1] && end > meeting[0]){
                overlaps.add(new int[]{Math.max(meeting[0], start), Math.min(end, meeting[1])});
            }
        }
        calendar.add(new int[]{start, end});
        return true;
    }
}