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:
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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.bookper test case will be at most1000. - In calls to
MyCalendar.book(start, end),startandendare integers in the range[0, 10^9].
Idea:
Brute Force
Solution1:
Brute Force
Time Complexity: O(n^2)
Space Complexity: O(n)
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// 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
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// 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
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""" 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)
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// 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_; }; |
Related Problems:
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