LeetCode #729 — MEDIUM

My Calendar I

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

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.

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 startTime and endTime that represents a booking on the half-open interval [startTime, endTime), the range of real numbers x such that startTime <= x < endTime.

Implement the MyCalendar class:

  • MyCalendar() Initializes the calendar object.
  • boolean book(int startTime, int endTime) Returns true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false and do not add the event to the calendar.

Example 1:

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.

Constraints:

  • 0 <= start < end <= 109
  • At most 1000 calls will be made to book.
Patterns Used
🔍Modified Binary Search→🏗Design Patterns→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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 startTime and endTime that represents a booking on the half-open interval [startTime, endTime), the range of real numbers x such that startTime <= x < endTime. Implement the MyCalendar class: MyCalendar() Initializes the calendar object. boolean book(int startTime, int endTime) Returns true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false and do not add the event to the calendar.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Design · Segment Tree

Example 1

["MyCalendar","book","book","book"]
[[],[10,20],[15,25],[20,30]]

Related Problems

  • My Calendar II (my-calendar-ii)
  • My Calendar III (my-calendar-iii)
  • Determine if Two Events Have Conflict (determine-if-two-events-have-conflict)
Step 02

Core Insight

What unlocks the optimal approach

  • Store the events as a sorted list of intervals. If none of the events conflict, then the new event can be added.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #729: My Calendar I
class MyCalendar {
    private final TreeMap<Integer, Integer> tm = new TreeMap<>();

    public MyCalendar() {
    }

    public boolean book(int startTime, int endTime) {
        var e = tm.ceilingEntry(startTime + 1);
        if (e != null && e.getValue() < endTime) {
            return false;
        }
        tm.put(endTime, startTime);
        return true;
    }
}

/**
 * Your MyCalendar object will be instantiated and called as such:
 * MyCalendar obj = new MyCalendar();
 * boolean param_1 = obj.book(startTime,endTime);
 */
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × log n)
Space
O(n)

Approach Breakdown

LINEAR SCAN
O(n) time
O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH
O(log n) time
O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Related Problems
#731My Calendar IImedium→#1157Online Majority Element In Subarrayhard→#2080Range Frequency Queriesmedium→#3508Implement Routermedium→#307Range Sum Query - Mutablemedium→