Problem

A rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) are the coordinates of its bottom-left corner, and (x2, y2) are the coordinates of its top-right corner.

Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.

Given two (axis-aligned) rectangles, return whether they overlap.

Example 1:

Input: rec1 = [0,0,2,2], rec2 = [1,1,3,3]
Output: true

Example 2:

Input: rec1 = [0,0,1,1], rec2 = [1,0,2,1]
Output: false

Notes:

Both rectangles rec1 and rec2 are lists of 4 integers.
All coordinates in rectangles will be between -10^9 and 10^9.

Solution: Geometry

Time complexity: O(1)

Space complexity: O(1)

// Author: Huahua

// Running time: 0 ms

class Solution {

public:

bool isRectangleOverlap(vector<int>& rec1, vector<int>& rec2) {

return rec1[0] < rec2[2] && rec2[0] < rec1[2] &&

rec1[1] < rec2[3] && rec2[1] < rec1[3];

}

};

Problem:

Implement a MyCalendarTwo class to store your events. A new event can be added if adding the event will not cause a triple booking.

Your class will have one method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.

A triple booking happens when three events have some non-empty intersection (ie., there is some time that is common to all 3 events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return false and do not add the event to the calendar.

Your class will be called like this: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)

Example 1:

MyCalendar();

MyCalendar.book(10, 20); // returns true

MyCalendar.book(50, 60); // returns true

MyCalendar.book(10, 40); // returns true

MyCalendar.book(5, 15); // returns false

MyCalendar.book(5, 10); // returns true

MyCalendar.book(25, 55); // returns true

Explanation<b>:</b>

The first two events can be booked. The third event can be double booked.

The fourth event (5, 15) can't be booked, because it would result in a triple booking.

The fifth event (5, 10) can be booked, as it does not use time 10 which is already double booked.

The sixth event (25, 55) can be booked, as the time in [25, 40) will be double booked with the third event;

the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

Note:

The number of calls to MyCalendar.book per test case will be at most 1000.
In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

Idea:

Brute Force

Solution1:

Brute Force

Time Complexity: O(n^2)

Space Complexity: O(n)

// Author: Huahua

// Runtime: 82 ms

class MyCalendarTwo {

public:

MyCalendarTwo() {}

bool book(int start, int end) {

for (const auto& kv : overlaps_)

if (max(start, kv.first) < min(end, kv.second)) return false;

for (const auto& kv : booked_) {

const int ss = max(start, kv.first);

const int ee = min(end, kv.second);

if (ss < ee) overlaps_.emplace_back(ss, ee);

}

booked_.emplace_back(start, end);

return true;

}

private:

vector<pair<int, int>> booked_;

vector<pair<int, int>> overlaps_;

};

Java

// Author: Huahua

// Runtime: 156 ms

class MyCalendarTwo {

private List<int[]> booked_;

private List<int[]> overlaps_;

public MyCalendarTwo() {

booked_ = new ArrayList<>();

overlaps_ = new ArrayList<>();

}

public boolean book(int start, int end) {

for (int[] range : overlaps_)

if (Math.max(range[0], start) < Math.min(range[1], end))

return false;

for (int[] range : booked_) {

int ss = Math.max(range[0], start);

int ee = Math.min(range[1], end);

if (ss < ee) overlaps_.add(new int[]{ss, ee});

}

booked_.add(new int[]{start, end});

return true;

}

Python

"""

Author: Huahua

Runtime: 638 ms

"""

class MyCalendarTwo:

def __init__(self):

self.booked_ = []

self.overlaps_ = []

def book(self, start, end):

for s, e in self.overlaps_:

if start < e and end > s: return False

for s, e in self.booked_:

if start < e and end > s:

self.overlaps_.append([max(start, s), min(end, e)])

self.booked_.append([start, end])

return True

Solution 2:

Counting

Time Complexity: O(n^2logn)

Space Complexity: O(n)

// Author: Huahua

// Runtime: 212 ms

class MyCalendarTwo {

public:

MyCalendarTwo() {}

bool book(int start, int end) {

++delta_[start];

--delta_[end];

int count = 0;

for (const auto& kv : delta_) {

count += kv.second;

if (count == 3) {

--delta_[start];

++delta_[end];

return false;

}

if (kv.first > end) break;

}

return true;

}

private:

map<int, int> delta_;

};

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