Last updated on July 19th, 2024 at 10:48 pm

Here, We see My Calendar II LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Breadth-First Search, Depth-First Search, Tree

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LinkedIn

Level of Question

Medium

My Calendar II LeetCode Solution

Problem Statement

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.

1. My Calendar II LeetCode Solution C++

class MyCalendarTwo {
public:
    map<int, int> m;
    MyCalendarTwo() {  
    }
    bool book(int start, int end) {
        int sum = 0;
        m[start]++;
        m[end]--;
        map<int,int>::iterator i;
        for (i=m.begin();i != m.end();i++) {
            sum = sum + i->second;
            if (sum >= 3) {
                m[start]--;
                m[end]++;
                return false;
            }
        } 
        return true;        
    }
};

2. My Calendar II LeetCode Solution Java

class MyCalendarTwo {
    Map<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 sum=0;
        for(int val : map.values()){
            sum += val;
            if(sum >= 3){
                map.put(start, map.get(start)-1);
                map.put(end, map.get(end)+1);
                if(map.get(start) == 0)
                    map.remove(start);
                return false;
            }
        }
        return true;
    }
}

3. My Calendar II LeetCode Solution JavaScript

var MyCalendarTwo = function() {
  this.calendar = [];
  this.overlaps = [];
};
MyCalendarTwo.prototype.book = function(start, end) {
  for (let date of this.overlaps) {
    if (start < date[1] && end > date[0]) return false;
  }
  for (let date of this.calendar) {
    if (start < date[1] && end > date[0]) {
      this.overlaps.push([Math.max(date[0], start), Math.min(date[1], end)]);
    }
  }
  this.calendar.push([start, end]); 
  return true;
};

4. My Calendar II LeetCode Solution Python

class MyCalendarTwo(object):
    def __init__(self):
        self.lst = []  
    def book(self, start, end):
        self.lst.append((start, +1))
        self.lst.append((end, -1))
        self.lst.sort()
        overlaps = 0
        for book in self.lst:
            overlaps += book[1]
            if overlaps > 2:
                self.lst.remove((start, +1))
                self.lst.remove((end, -1))
                return False
        return True