Problem

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.

Examples

Example 1:

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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.

Solution

Method 1 - Using TreeMap and Count

Visualize this process as “scanning” from left to right with a “vertical laser”. Each endpoint (either a starting point or an ending point) is treated as an event—where a starting point increments by +1 and an ending point decrements by -1. The cumulative value “count” represents the number of “active” intervals intersected by the vertical laser.

Code

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class MyCalendarTwo {

    private final TreeMap<Integer, Integer> map;    
    
    public MyCalendarTwo() {
        map = new TreeMap<>();
    }
    
    public boolean book(int start, int end) {
        map.put(start, map.getOrDefault(start, 0) + 1);
        map.put(end, map.getOrDefault(end, 0) - 1);
        int count = 0;
        for(Map.Entry<Integer, Integer> entry : map.entrySet()) {
            count += entry.getValue();
            if(count > 2) {
                map.put(start, map.get(start) - 1);
                if(map.get(start) == 0) {
                    map.remove(start);
                }
                map.put(end, map.get(end) + 1);
                if(map.get(end) == 0) {
                    map.remove(end);
                }
                return false;
            }
        }
        return true;
    }
}

Dry Run

Lets take input for book as:

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bookings = [[10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]

book=[10,20]
map = {10:1, 20:-1}
count = 1+-1  = 0
output = true

book = [50,60]
map = {10:1, 20:-1, 50:1, 60:-1}
count = 1+-1+1+-1
ouptut = true

book=[10,40]
map = {10:2, 20:-1, 40:-1, 50:1, 60:-1}
count = 2 + -1 + -1 + 1+ -1 = 0 ; so Count never exceeds 2
output = true

book = [5,15]
map = {5: 1, 10:2, 15: -1, 20:-1, 40:-1, 50:1, 60:-1}
count = 1+2 ... count exceeds 2, so we cannot add this value
output = false
But because we added the interval earlier, we should remove it as we are returning false.

Observe how the map’s ordering shifts during a dry run. This is why we are utilising a TreeMap. Refer: Java TreeMap Class