LeetCode Problem
731. My Calendar II
Link to LeetCode
You are implementing a program to use as your calendar. We can add a new event if adding the event will not cause a triple booking.
A triple booking happens when three events have some non-empty intersection (i.e., some moment is common to all the three 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 MyCalendarTwo class:
MyCalendarTwo() 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 triple booking. Otherwise, return false and do not add the event to the calendar.
Input
["MyCalendarTwo", "book", "book", "book", "book", "book", "book"]
[[], [10, 20], [50, 60], [10, 40], [5, 15], [5, 10], [25, 55]]
Output
[null, true, true, true, false, true, true]
Explanation
MyCalendarTwo myCalendarTwo = new MyCalendarTwo();
myCalendarTwo.book(10, 20); // return True, The event can be booked.
myCalendarTwo.book(50, 60); // return True, The event can be booked.
myCalendarTwo.book(10, 40); // return True, The event can be double booked.
myCalendarTwo.book(5, 15); // return False, The event cannot be booked, because it would result in a triple booking.
myCalendarTwo.book(5, 10); // return True, The event can be booked, as it does not use time 10 which is already double booked.
myCalendarTwo.book(25, 55); // return True, The event 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.
Analysis
This is a more complicated version of My Calendar I, in which we only care if a time interval is used already. For this problem, we can add another list to track the double booked intervals.
Time complexity is O(N^2) as each book operation iterate over all existing intervals. Space complexity is O(N).
class MyCalendarTwo {
ArrayList single = null;
ArrayList overlap = null;
public MyCalendarTwo() {
single = new ArrayList<>();
overlap = new ArrayList<>();
}
public boolean book(int start, int end) {
for (int[] itv : overlap) {
if (end > itv[0] && start < itv[1]) {
return false;
}
}
for (int[] itv : single) {
if (end > itv[0] && start < itv[1]) {
overlap.add(new int[]{Math.max(itv[0], start), Math.min(itv[1], end)});
}
}
single.add(new int[]{start, end});
return true;
}
}