LeetCode #732 — HARD

My Calendar III

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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.
int book(int startTime, int endTime) Returns an integer k representing the largest integer such that there exists a k-booking in the calendar.

Example 1:

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

Constraints:

0 <= startTime < endTime <= 10⁹
At most 400 calls will be made to book.

Step 01

Brute Force Baseline

Problem summary: 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. int book(int startTime, int endTime) Returns an integer k representing the largest integer such that there exists a k-booking in the calendar.

Baseline thinking

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

Pattern signal: Binary Search · Design · Segment Tree

Example 1

["MyCalendarThree","book","book","book","book","book","book"]
[[],[10,20],[50,60],[10,40],[5,15],[5,10],[25,55]]

Core Insight

What unlocks the optimal approach

Treat each interval [start, end) as two events "start" and "end", and process them in sorted order.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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

Largest constraint values

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 #732: My Calendar III
class Node {
    Node left;
    Node right;
    int l;
    int r;
    int mid;
    int v;
    int add;
    public Node(int l, int r) {
        this.l = l;
        this.r = r;
        this.mid = (l + r) >> 1;
    }
}

class SegmentTree {
    private Node root = new Node(1, (int) 1e9 + 1);

    public SegmentTree() {
    }

    public void modify(int l, int r, int v) {
        modify(l, r, v, root);
    }

    public void modify(int l, int r, int v, Node node) {
        if (l > r) {
            return;
        }
        if (node.l >= l && node.r <= r) {
            node.v += v;
            node.add += v;
            return;
        }
        pushdown(node);
        if (l <= node.mid) {
            modify(l, r, v, node.left);
        }
        if (r > node.mid) {
            modify(l, r, v, node.right);
        }
        pushup(node);
    }

    public int query(int l, int r) {
        return query(l, r, root);
    }

    public int query(int l, int r, Node node) {
        if (l > r) {
            return 0;
        }
        if (node.l >= l && node.r <= r) {
            return node.v;
        }
        pushdown(node);
        int v = 0;
        if (l <= node.mid) {
            v = Math.max(v, query(l, r, node.left));
        }
        if (r > node.mid) {
            v = Math.max(v, query(l, r, node.right));
        }
        return v;
    }

    public void pushup(Node node) {
        node.v = Math.max(node.left.v, node.right.v);
    }

    public void pushdown(Node node) {
        if (node.left == null) {
            node.left = new Node(node.l, node.mid);
        }
        if (node.right == null) {
            node.right = new Node(node.mid + 1, node.r);
        }
        if (node.add != 0) {
            Node left = node.left, right = node.right;
            left.add += node.add;
            right.add += node.add;
            left.v += node.add;
            right.v += node.add;
            node.add = 0;
        }
    }
}

class MyCalendarThree {
    private SegmentTree tree = new SegmentTree();

    public MyCalendarThree() {
    }

    public int book(int start, int end) {
        tree.modify(start + 1, end, 1);
        return tree.query(1, (int) 1e9 + 1);
    }
}

/**
 * Your MyCalendarThree object will be instantiated and called as such:
 * MyCalendarThree obj = new MyCalendarThree();
 * int param_1 = obj.book(start,end);
 */

// Accepted solution for LeetCode #732: My Calendar III
type node struct {
	left      *node
	right     *node
	l, mid, r int
	v, add    int
}

func newNode(l, r int) *node {
	return &node{
		l:   l,
		r:   r,
		mid: int(uint(l+r) >> 1),
	}
}

type segmentTree struct {
	root *node
}

func newSegmentTree() *segmentTree {
	return &segmentTree{
		root: newNode(1, 1e9+1),
	}
}

func (t *segmentTree) modify(l, r, v int, n *node) {
	if l > r {
		return
	}
	if n.l >= l && n.r <= r {
		n.v += v
		n.add += v
		return
	}
	t.pushdown(n)
	if l <= n.mid {
		t.modify(l, r, v, n.left)
	}
	if r > n.mid {
		t.modify(l, r, v, n.right)
	}
	t.pushup(n)
}

func (t *segmentTree) query(l, r int, n *node) int {
	if l > r {
		return 0
	}
	if n.l >= l && n.r <= r {
		return n.v
	}
	t.pushdown(n)
	v := 0
	if l <= n.mid {
		v = max(v, t.query(l, r, n.left))
	}
	if r > n.mid {
		v = max(v, t.query(l, r, n.right))
	}
	return v
}

func (t *segmentTree) pushup(n *node) {
	n.v = max(n.left.v, n.right.v)
}

func (t *segmentTree) pushdown(n *node) {
	if n.left == nil {
		n.left = newNode(n.l, n.mid)
	}
	if n.right == nil {
		n.right = newNode(n.mid+1, n.r)
	}
	if n.add != 0 {
		n.left.add += n.add
		n.right.add += n.add
		n.left.v += n.add
		n.right.v += n.add
		n.add = 0
	}
}

type MyCalendarThree struct {
	tree *segmentTree
}

func Constructor() MyCalendarThree {
	return MyCalendarThree{newSegmentTree()}
}

func (this *MyCalendarThree) Book(start int, end int) int {
	this.tree.modify(start+1, end, 1, this.tree.root)
	return this.tree.query(1, int(1e9)+1, this.tree.root)
}

/**
 * Your MyCalendarThree object will be instantiated and called as such:
 * obj := Constructor();
 * param_1 := obj.Book(start,end);
 */

# Accepted solution for LeetCode #732: My Calendar III
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)

// Accepted solution for LeetCode #732: My Calendar III
use std::collections::BTreeMap;

struct MyCalendarThree {
    data: BTreeMap<i32, i32>,
}

impl MyCalendarThree {
    fn new() -> Self {
        let data = BTreeMap::new();
        MyCalendarThree { data }
    }

    fn book(&mut self, start: i32, end: i32) -> i32 {
        *self.data.entry(start).or_default() += 1;
        *self.data.entry(end).or_default() -= 1;
        let mut cur = 0;
        let mut res = 0;
        for v in self.data.values() {
            cur += v;
            res = res.max(cur);
        }
        res
    }
}

#[test]
fn test() {
    let mut obj = MyCalendarThree::new();
    assert_eq!(obj.book(10, 20), 1);
    assert_eq!(obj.book(50, 60), 1);
    assert_eq!(obj.book(10, 40), 2);
    assert_eq!(obj.book(5, 15), 3);
    assert_eq!(obj.book(5, 10), 3);
    assert_eq!(obj.book(25, 55), 3);
}

// Accepted solution for LeetCode #732: My Calendar III
class Node {
    left: Node | null = null;
    right: Node | null = null;
    l: number;
    r: number;
    mid: number;
    v: number = 0;
    add: number = 0;

    constructor(l: number, r: number) {
        this.l = l;
        this.r = r;
        this.mid = (l + r) >> 1;
    }
}

class SegmentTree {
    private root: Node = new Node(1, 1e9 + 1);

    constructor() {}

    modify(l: number, r: number, v: number, node: Node = this.root): void {
        if (l > r) {
            return;
        }
        if (node.l >= l && node.r <= r) {
            node.v += v;
            node.add += v;
            return;
        }
        this.pushdown(node);
        if (l <= node.mid) {
            this.modify(l, r, v, node.left!);
        }
        if (r > node.mid) {
            this.modify(l, r, v, node.right!);
        }
        this.pushup(node);
    }

    query(l: number, r: number, node: Node = this.root): number {
        if (l > r) {
            return 0;
        }
        if (node.l >= l && node.r <= r) {
            return node.v;
        }
        this.pushdown(node);
        let v = 0;
        if (l <= node.mid) {
            v = Math.max(v, this.query(l, r, node.left!));
        }
        if (r > node.mid) {
            v = Math.max(v, this.query(l, r, node.right!));
        }
        return v;
    }

    private pushup(node: Node): void {
        node.v = Math.max(node.left!.v, node.right!.v);
    }

    private pushdown(node: Node): void {
        if (node.left === null) {
            node.left = new Node(node.l, node.mid);
        }
        if (node.right === null) {
            node.right = new Node(node.mid + 1, node.r);
        }
        if (node.add !== 0) {
            const left = node.left!;
            const right = node.right!;
            left.add += node.add;
            right.add += node.add;
            left.v += node.add;
            right.v += node.add;
            node.add = 0;
        }
    }
}

class MyCalendarThree {
    private tree: SegmentTree;

    constructor() {
        this.tree = new SegmentTree();
    }

    book(start: number, end: number): number {
        this.tree.modify(start + 1, end, 1);
        return this.tree.query(1, 1e9 + 1);
    }
}

/**
 * Your MyCalendarThree object will be instantiated and called as such:
 * var obj = new MyCalendarThree()
 * var param_1 = obj.book(startTime, endTime)
 */

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

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.

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.