grid47 Exploring patterns and algorithms
Aug 26, 2024
5 min read
Solution to LeetCode 729: My Calendar I Problem
You are implementing a calendar application where you can add events. Each event has a start time and end time represented by a half-open interval [start, end). An event can only be added if it does not overlap with existing events in the calendar. Return true if the event can be booked successfully, and false if it causes a conflict.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of method calls with parameters. The first call is always initializing the MyCalendar object, followed by calls to the book method.
Output: The output will be an array of results for each call. For the MyCalendar initialization, the result will be null, and for each book method call, the result will be a boolean indicating whether the event was successfully booked or not.
Example: Output: [null, true, false, true]
Constraints:
• For each call to book, return true if the event can be booked, false otherwise.
Goal: The goal is to implement a calendar that checks for overlapping events and returns whether a new event can be added without causing a conflict.
Steps:
• 1. Create a map or structure to store the events, with start times as the key.
• 2. For each new event, check if it overlaps with any existing event.
• 3. If no overlap is found, add the event to the calendar and return true.
• 4. If overlap is found, return false and do not add the event.
Goal: The constraints define the input limits.
Steps:
• 0 <= start < end <= 10^9
• 1 <= k <= 1000
Assumptions:
• The input start and end times are integers, and the events are well-formed.
• Explanation: In the first example, the first event is successfully booked as no previous events exist. The second event cannot be booked as it overlaps with the first event. The third event is successfully booked since it does not overlap with the first event (ends at 20).
Approach: The approach involves maintaining a data structure to track events and efficiently checking for overlaps when booking a new event.
Observations:
• We need to efficiently store and check for overlapping intervals.
• Using a map structure will help track the intervals and quickly check for any conflicts.
Steps:
• 1. Use a map to store events by end time.
• 2. For each new booking, check if there is an existing event with a conflicting time range.
• 3. If no conflict, add the new event and return true; otherwise, return false.
Empty Inputs:
• If no events have been booked yet, any new event can be booked.
Large Inputs:
• Ensure the solution handles up to 1000 booking attempts efficiently.
Special Values:
• Consider cases where the start and end times are very close, or the events have minimal overlap.
Constraints:
• The solution must be able to handle start and end times within the range of [0, 10^9].
public:MyCalendar() {
}
boolbook(int start, int end) {
auto nxt = mp.upper_bound(start);
if (nxt != mp.end() && (*nxt).second < end) {
returnfalse;
}
mp[end] = start;
returntrue;
}
};
/**
* Your MyCalendar object will be instantiated and called as such:
* MyCalendar* obj = new MyCalendar();
* bool param_1 = obj->book(start,end);
1 : Access Modifier
public:
Defines the access level for the members that follow, making them accessible to other parts of the program.
2 : Constructor
MyCalendar() {
Constructor for the MyCalendar class. Initializes the calendar object.
3 : Method Definition
boolbook(int start, int end) {
Defines the 'book' method which attempts to schedule a new booking from 'start' to 'end'.
4 : Variable Declaration
auto nxt = mp.upper_bound(start);
Uses 'upper_bound' to find the first event in 'mp' that starts after the 'start' time.
5 : Conditional Check
if (nxt != mp.end() && (*nxt).second < end) {
Checks if there is an existing booking that overlaps with the new booking.
6 : Return Statement
returnfalse;
If an overlap is detected, return false to indicate the booking was not successful.
7 : Map Update
mp[end] = start;
Records the new booking in the map 'mp', where 'end' is the key and 'start' is the value.
8 : Return Statement
returntrue;
If no overlap is found, return true to indicate the booking was successful.
Best Case: O(log N)
Average Case: O(log N)
Worst Case: O(N)
Description: In the worst case, we check all the events for conflicts, making the time complexity O(N), where N is the number of events. A more efficient approach could be implemented using a balanced tree or interval tree for quicker access.
Best Case: O(N)
Worst Case: O(N)
Description: The space complexity is O(N) as we store all booked events in the calendar.
