Leetcode #729: My Calendar I
In this guide, we solve Leetcode #729 My Calendar I 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 double booking.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Segment Tree, Array, Binary Search, Ordered Set
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
["MyCalendar", "book", "book", "book"]
[[], [10, 20], [15, 25], [20, 30]]
Output
[null, true, false, true]
Explanation
MyCalendar myCalendar = new MyCalendar();
myCalendar.book(10, 20); // return True
myCalendar.book(15, 25); // return False, It can not be booked because time 15 is already booked by another event.
myCalendar.book(20, 30); // return True, The event can be booked, as the first event takes every time less than 20, but not including 20.
Python Solution
class MyCalendar:
def __init__(self):
self.sd = SortedDict()
def book(self, start: int, end: int) -> bool:
idx = self.sd.bisect_right(start)
if idx < len(self.sd) and self.sd.values()[idx] < end:
return False
self.sd[end] = start
return True
# Your MyCalendar object will be instantiated and called as such:
# obj = MyCalendar()
# param_1 = obj.book(start,end)
Complexity
The time complexity is , and the space complexity is . The space complexity is .
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.