MyCalendarI.md

The question

You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a double booking.

A double booking happens when two events have some non-empty intersection (i.e., some moment is common to both 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 MyCalendar class:

MyCalendar() 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 double booking. Otherwise, return false and do not add the event to the calendar.

Example 1:

Input
["MyCalendar", "book", "book", "book"]
[[], [10, 20], [15, 25], [20, 30]]
Output
[null, true, false, true]

Explanation
MyCalendar myCalendar = new MyCalendar();
myCalendar.book(10, 20); // return True
myCalendar.book(15, 25); // return False, It can not be booked because time 15 is already booked by another event.
myCalendar.book(20, 30); // return True, The event can be booked, as the first event takes every time less than 20, but not including 20.

Approach:- First thought of using simple for loop and check each event entry, but time complexity shot up to O(n^2). So how can this be reduced. Introducing segment trees. Why? Because they give O(logN) time complexity for searching elements!!

Lets code

    TreeMap<Integer, Integer> calendar;

    MyCalendar() {
        calendar = new TreeMap();
    }

    public boolean book(int start, int end) {
      Integer prev = calendar.floorKey(start);
              next = calender.ceilingKey(start);
              
      if((prev==null || calender.get(prev)<=start) && (next==null || end <= next)){
          calender.add(start, end);
          return true;
      }
      
      return false;
    }

Time complexity - O(NlogN) Space complexity - O(N)