# Solving Leetcode Interviews in Seconds with AI: My Calendar III


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "732" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> A k-booking happens when k events have some non-empty intersection (i.e., there is some time that is common to all k events.) You are given some events [startTime, endTime), after each given event, return an integer k representing the maximum k-booking between all the previous events. Implement the MyCalendarThree class:  MyCalendarThree() Initializes the object. int book(int startTime, int endTime) Returns an integer k representing the largest integer such that there exists a k-booking in the calendar.    Example 1:  Input ["MyCalendarThree", "book", "book", "book", "book", "book", "book"] [[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]] Output [null, 1, 1, 2, 3, 3, 3]  Explanation MyCalendarThree myCalendarThree = new MyCalendarThree(); myCalendarThree.book(10, 20); // return 1 myCalendarThree.book(50, 60); // return 1 myCalendarThree.book(10, 40); // return 2 myCalendarThree.book(5, 15); // return 3 myCalendarThree.book(5, 10); // return 3 myCalendarThree.book(25, 55); // return 3     Constraints:  0 <= startTime < endTime <= 109 At most 400 calls will be made to book.  

	# Explanation
	Here's a breakdown of the approach, followed by the Python code:

*   **Difference Arrays/Sweep Line:** The core idea is to use a "difference array" technique (similar to a sweep line algorithm). We represent the calendar as a series of start and end points. For each event, we increment the count at the start time and decrement the count at the end time.

*   **Tracking Overlaps:** By iterating through the sorted timestamps, we keep a running count of active events. The maximum value of this running count represents the maximum k-booking.

*   **Sorted Data:** Using a sorted data structure (like a dictionary) maintains the order of timestamps, which is crucial for the sweep line approach.

*   **Runtime & Storage Complexity:**
    *   Runtime: O(NlogN) where N is the number of bookings, due to sorting the timestamps.
    *   Storage: O(N) to store the calendar events.

	
	# Code
	```python
	class MyCalendarThree:

    def __init__(self):
        self.calendar = {}

    def book(self, startTime: int, endTime: int) -> int:
        self.calendar[startTime] = self.calendar.get(startTime, 0) + 1
        self.calendar[endTime] = self.calendar.get(endTime, 0) - 1

        sorted_times = sorted(self.calendar.keys())
        active_events = 0
        max_bookings = 0

        for time in sorted_times:
            active_events += self.calendar[time]
            max_bookings = max(max_bookings, active_events)

        return max_bookings
	```