Leetcode #732: My Calendar III
In this guide, we solve Leetcode #732 My Calendar III 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
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Segment Tree, 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
["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
Python Solution
class Node:
def __init__(self, l, r):
self.left = None
self.right = None
self.l = l
self.r = r
self.mid = (l + r) >> 1
self.v = 0
self.add = 0
class SegmentTree:
def __init__(self):
self.root = Node(1, int(1e9 + 1))
def modify(self, l: int, r: int, v: int, node: Node = None):
if l > r:
return
if node is None:
node = self.root
if node.l >= l and node.r <= r:
node.v += v
node.add += v
return
self.pushdown(node)
if l <= node.mid:
self.modify(l, r, v, node.left)
if r > node.mid:
self.modify(l, r, v, node.right)
self.pushup(node)
def query(self, l: int, r: int, node: Node = None) -> int:
if l > r:
return 0
if node is None:
node = self.root
if node.l >= l and node.r <= r:
return node.v
self.pushdown(node)
v = 0
if l <= node.mid:
v = max(v, self.query(l, r, node.left))
if r > node.mid:
v = max(v, self.query(l, r, node.right))
return v
def pushup(self, node: Node):
node.v = max(node.left.v, node.right.v)
def pushdown(self, node: Node):
if node.left is None:
node.left = Node(node.l, node.mid)
if node.right is None:
node.right = Node(node.mid + 1, node.r)
if node.add:
node.left.v += node.add
node.right.v += node.add
node.left.add += node.add
node.right.add += node.add
node.add = 0
class MyCalendarThree:
def __init__(self):
self.tree = SegmentTree()
def book(self, start: int, end: int) -> int:
self.tree.modify(start + 1, end, 1)
return self.tree.query(1, int(1e9 + 1))
# Your MyCalendarThree object will be instantiated and called as such:
# obj = MyCalendarThree()
# param_1 = obj.book(start,end)
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.