花花酱 LeetCode 732. My Calendar III

zxi — Thu, 07 Dec 2017 05:29:41 +0000

Problem:

link: https://leetcode.com/problems/my-calendar-iii/description/

Implement a MyCalendarThree class to store your events. A new event can always be added.

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

For each call to the method MyCalendar.book, return an integer K representing the largest integer such that there exists a K-booking in the calendar.

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

Example 1:

MyCalendarThree();
MyCalendarThree.book(10, 20); // returns 1
MyCalendarThree.book(50, 60); // returns 1
MyCalendarThree.book(10, 40); // returns 2
MyCalendarThree.book(5, 15); // returns 3
MyCalendarThree.book(5, 10); // returns 3
MyCalendarThree.book(25, 55); // returns 3
Explanation: 
The first two events can be booked and are disjoint, so the maximum K-booking is a 1-booking.
The third event [10, 40) intersects the first event, and the maximum K-booking is a 2-booking.
The remaining events cause the maximum K-booking to be only a 3-booking.
Note that the last event locally causes a 2-booking, but the answer is still 3 because
eg. [10, 20), [10, 40), and [5, 15) are still triple booked.

Note:

The number of calls to MyCalendarThree.book per test case will be at most 400.
In calls to MyCalendarThree.book(start, end), start and end are integers in the range [0, 10^9].

Idea:

Similar to LeetCode 731 My Calendar II Use an ordered / tree map to track the number of event at current time.

For a new book event, increase the number of events at start, decrease the number of events at end.

Scan the timeline to find the maximum number of events.

Solution 1: Count Boundaries

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua
// Runtime: 116 ms
class MyCalendarThree {
public:
    MyCalendarThree() {}
    
    int book(int start, int end) {
        ++deltas_[start];
        --deltas_[end];
        int ans = 0;
        int curr = 0;
        for (const auto& kv : deltas_)            
            ans = max(ans, curr += kv.second);
        return ans;
    }
private:
    map deltas_;
};

Solution 2

C++

// Author: Huahua
// Runtime: 66 ms
class MyCalendarThree {
public:
    MyCalendarThree() {
        counts_[INT_MIN] = 0;
        counts_[INT_MAX] = 0;
        max_count_ = 0;
    }
    
    int book(int start, int end) {        
        auto l = prev(counts_.upper_bound(start));   // l->first < start
        auto r = counts_.lower_bound(end);           // r->first >= end
        for (auto curr = l, next = curr; curr != r; curr = next) {
            ++next;
            
            if (next->first > end)
                counts_[end] = curr->second;
            
            if (curr->first <= start && next->first > start)
                max_count_ = max(max_count_, counts_[start] = curr->second + 1);            
            else
                max_count_ = max(max_count_, ++curr->second);
        }
        
        return max_count_;
    }
private:
    map counts_;
    int max_count_;
};

Solution 3: Segment Tree

C++

// Author: Huahua
// Runtime: 63 ms (<95.65%)
class MyCalendarThree {
public:
    MyCalendarThree(): max_count_(0) {
        root_.reset(new Node(0, 100000000, 0));        
    }
    
    int book(int start, int end) {
        Add(start, end, root_.get());
        return max_count_;
    }
private:
    struct Node {
        Node(int l, int r, int count):l(l), m(-1), r(r), count(count){}
        int l;
        int m;
        int r;
        int count;
        std::unique_ptr left;
        std::unique_ptr right;
    };
    
    void Add(int start, int end, Node* root) {
        if (root->m != -1) {
            if (end <= root->m) 
                Add(start, end, root->left.get());
            else if(start >= root->m)
                Add(start, end, root->right.get());
            else {
                Add(start, root->m, root->left.get());
                Add(root->m, end, root->right.get());
            }
            return;
        }
        
        if (start == root->l && end == root->r)
            max_count_ = max(max_count_, ++root->count);
        else if (start == root->l) {
            root->m = end;
            root->left.reset(new Node(start, root->m, root->count + 1));
            root->right.reset(new Node(root->m, root->r, root->count));
            max_count_ = max(max_count_, root->count + 1);
        } else if(end == root->r) {
            root->m = start;
            root->left.reset(new Node(root->l, root->m, root->count));
            root->right.reset(new Node(root->m, root->r, root->count + 1));
            max_count_ = max(max_count_, root->count + 1);
        } else {
            root->m = start;
            root->left.reset(new Node(root->l, root->m, root->count));
            root->right.reset(new Node(root->m, root->r, root->count));
            Add(start, end, root->right.get());
        }
    }
    
    std::unique_ptr root_;
    int max_count_;
};

Python3

"""
Author: Huahua
Runtime: 436 ms (<88.88%)
"""
class Node:
    def __init__(self, l, r, count):
        self.l = l
        self.m = -1
        self.r = r            
        self.count = count
        self.left = None
        self.right = None
            
class MyCalendarThree:        
    def __init__(self):
        self.root = Node(0, 10**9, 0)
        self.max = 0

    def book(self, start, end):
        self.add(start, end, self.root)
        return self.max
    
    def add(self, start, end, root):
        if root.m != -1:
            if end <= root.m: self.add(start, end, root.left)
            elif start >= root.m: self.add(start, end, root.right)
            else:
                self.add(start, root.m, root.left)
                self.add(root.m, end, root.right)
            return
        
        if start == root.l and end == root.r:
            root.count += 1
            self.max = max(self.max, root.count)
        elif start == root.l:
            root.m = end
            root.left = Node(start, root.m, root.count + 1)
            root.right = Node(root.m, root.r, root.count)
            self.max = max(self.max, root.count + 1)
        elif end == root.r:
            root.m = start;
            root.left = Node(root.l, root.m, root.count)
            root.right = Node(root.m, root.r, root.count + 1)
            self.max = max(self.max, root.count + 1)
        else:
            root.m = start
            root.left = Node(root.l, root.m, root.count)
            root.right = Node(root.m, root.r, root.count)
            self.add(start, end, root.right)

Related Problems:

The post 花花酱 LeetCode 732. My Calendar III appeared first on Huahua's Tech Road.

ordered map Archives - Huahua's Tech Road

花花酱 LeetCode 732. My Calendar III

Solution 1: Count Boundaries

C++

Solution 2

C++

Solution 3: Segment Tree

C++

Python3