LeetCode My Calendar I Solution
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
end that represents a booking on the half-open interval
[start, end), the range of real numbers
x such that
start <= x < end.
MyCalendar()Initializes the calendar object.
boolean book(int start, int end)Returns
trueif the event can be added to the calendar successfully without causing a double booking. Otherwise, return
falseand do not add the event to the calendar.
Input ["MyCalendar", "book", "book", "book"] [, [10, 20], [15, 25], [20, 30]] Output [null, true, false, true]
1MyCalendar myCalendar = new MyCalendar(); 2myCalendar.book(10, 20); // return True 3myCalendar.book(15, 25); // return False, It can not be booked because time 15 is already booked by another event. 4myCalendar.book(20, 30); // return True, The event can be booked, as the first event takes every time less than 20, but not including 20.
0 <= start < end <= 109
- At most
1000calls will be made to
Problem Link: https://leetcode.com/problems/my-calendar-i/
To implement the booking behaviour, we will use binary search to find a potential insertion index, then check whether the new booking can be
actually scheduled into our calendar by checking whether the new booking overlaps with
1class MyCalendar: 2 def __init__(self): 3 self.calendar =  4 5 def book(self, start: int, end: int) -> bool: 6 left, right, idx = 0, len(self.calendar)-1, len(self.calendar) 7 while left <= right: 8 mid = (left + right) // 2 9 if self.calendar[mid] > start: 10 idx = mid 11 right = mid - 1 12 else: 13 left = mid + 1 14 # check if calendar[idx-1] or calendar[idx] overlaps with start and end 15 if (idx > 0 and self.calendar[idx-1] > start) or (idx < len(self.calendar) and self.calendar[idx] < end): 16 return False 17 self.calendar.insert(idx, (start, end)) 18 return True