What is the brute force approach and complexity for My Calendar I?

Keep a list of bookings; scan all of them on every call with time complexity O(N) and space complexity O(N).

What is the optimal approach and complexity for My Calendar I?

map start → end; only the booking at lower_bound(start) and the one before it can clash with optimal time complexity O(log N) and space complexity O(N).

How do you identify the pattern for My Calendar I?

Look for key signals in the prompt: "book(start, end) returns true/false" indicates Online version of Meeting Rooms: events arrive one at a time, so we can't sort them all up front.; "no double booking" indicates At most 1 event at any moment → only the neighbours of the new event can clash.; "half-open [start, end)" indicates Touching ends are fine; the overlap test uses strict `<` on both sides..

My Calendar I

Medium·LeetCode #729 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

Build a calendar class that never allows a double booking: two events that share some moment of time.

MyCalendar() creates an empty calendar.
bool book(int startTime, int endTime) tries to add an event covering the half-open range [startTime, endTime), meaning every time t with startTime ≤ t < endTime. If it does not overlap any event already in the calendar, add it and return true. Otherwise do not add it and return false.

Because the end is not included, an event ending at 20 and one starting at 20 do not overlap.

Input

A sequence of book(startTime, endTime) calls.

Output

For each call, true if the event was booked, false if it was rejected.

Example 1

Input: book(10, 20), book(15, 25), book(20, 30)

Output: true, false, true

[10,20) is booked. [15,25) overlaps it at 15 … 19, so it is rejected. [20,30) starts exactly when [10,20) ends, so it fits.

Constraints

• 0 ≤ startTime < endTime ≤ 10^9
• At most 1000 calls to book

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“book(start, end) returns true/false”

→ Online version of Meeting Rooms: events arrive one at a time, so we can't sort them all up front.

“no double booking”

→ At most 1 event at any moment → only the neighbours of the new event can clash.

“half-open [start, end)”

→ Touching ends are fine; the overlap test uses strict `<` on both sides.

Approach Complexity Ladder

Brute Force — check every bookingO(N)O(N)

Keep a list of bookings; scan all of them on every call

Optimal — sorted map, check two neighboursO(log N)O(N)

map start → end; only the booking at lower_bound(start) and the one before it can clash

Brute Force Intuition

Store every accepted booking in a vector. For a new event [s, e), loop over all bookings [s2, e2) and reject if s < e2 && s2 < e. If none clash, push it into the vector.

Recognition rule for this sub-pattern:

How many things are active at the same time? Sweep events in time order, or keep a min-heap of end times.

Remember this · Intervals

Trigger

Input is [start, end] pairs; overlap, merge, insert, meeting rooms, free time, minimum removals.

Think

After sorting, do I only ever need to compare against the most recent interval?

Key question

Sort by start or by end — which one makes the greedy choice obvious?

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