Leetcode #731: My Calendar II
In this guide, we solve Leetcode #731 My Calendar II in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Segment Tree, Array, Binary Search, Ordered Set, Prefix Sum
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
class MyCalendarTwo:
def __init__(self):
self.sd = SortedDict()
def book(self, startTime: int, endTime: int) -> bool:
self.sd[startTime] = self.sd.get(startTime, 0) + 1
self.sd[endTime] = self.sd.get(endTime, 0) - 1
s = 0
for v in self.sd.values():
s += v
if s > 2:
self.sd[startTime] -= 1
self.sd[endTime] += 1
return False
return True
# Your MyCalendarTwo object will be instantiated and called as such:
# obj = MyCalendarTwo()
# param_1 = obj.book(startTime,endTime)
Complexity
The time complexity is , and the space complexity is , where is the number of bookings. The space complexity is , where is the number of bookings.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.