# 731. My Calendar II LeetCode Solution | Easy Approach

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My Calendar II 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.

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.

Constraints:

• `0 <= start < end <= 109`
• At most `1000` calls will be made to `book`.

Time: O(n)
Space: O(n)

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