729. My Calendar I

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 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]

Example 2:

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.

Solution (Java)

class MyCalendar {

    static class Meeting implements Comparable<Meeting> {
        public final int start;
        public final int end;

        public Meeting(int start, int end) {
            this.start = start;
            this.end = end;
        }

        @Override
        public int compareTo(Meeting anotherMeeting) {
            return this.start - anotherMeeting.start;
        }
    }

    private final TreeSet<Meeting> meetings;

    public MyCalendar() {
        meetings = new TreeSet<>();
    }

    public boolean book(int start, int end) {
        Meeting meetingToBook = new Meeting(start, end);
        Meeting prevMeeting = meetings.floor(meetingToBook);
        Meeting nextMeeting = meetings.ceiling(meetingToBook);
        if ((prevMeeting == null || prevMeeting.end <= meetingToBook.start)
                && (nextMeeting == null || meetingToBook.end <= nextMeeting.start)) {

            meetings.add(meetingToBook);
            return true;
        }
        return false;
    }
}

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

Solution (Javascript)

var MyCalendar = function() {
    this.root = null
};

/** 
 * @param {number} start 
 * @param {number} end
 * @return {boolean}
 */
MyCalendar.prototype.book = function(start, end) {
  if (!this.root) {
    this.root = new TreeNode(start, end)
    return true
  }
  return insert(this.root, new TreeNode(start, end))[1]
};

/** 
 * Your MyCalendar object will be instantiated and called as such:
 * var obj = new MyCalendar()
 * var param_1 = obj.book(start,end)
 */
const TreeNode = function (start, end) {
  this.start = start
  this.end = end
  this.left = null
  this.right = null
}

const insert = (node, newNode) => {
  if (node === null) {
    return [newNode, true]
  }
  if (newNode.start >= node.end) {
    const [right, valid] = insert(node.right, newNode)
    node.right = right
    return [node, valid]
  } if (newNode.end <= node.start) {
    const [left, valid] = insert(node.left, newNode)
    node.left = left
    return [node, valid]
  }
  return [node, false]
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).