732

732. My Calendar III (Hard)

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 [start, end), 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 start, int end) 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, The first event can be booked and is disjoint, so the maximum k-booking is a 1-booking.
myCalendarThree.book(50, 60); // return 1, The second event can be booked and is disjoint, so the maximum k-booking is a 1-booking.
myCalendarThree.book(10, 40); // return 2, The third event [10, 40) intersects the first event, and the maximum k-booking is a 2-booking.
myCalendarThree.book(5, 15); // return 3, The remaining events cause the maximum K-booking to be only a 3-booking.
myCalendarThree.book(5, 10); // return 3
myCalendarThree.book(25, 55); // return 3

 

Constraints:

• 0 <= start < end <= 109
• At most 400 calls will be made to book.

Companies:
Google

Related Topics:
Segment Tree, Ordered Map

Similar Questions:

• My Calendar I (Medium)
• My Calendar II (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/my-calendar-iii/
// Author: github.com/lzl124631x
// Time:
//      MyCalendarThree: O(1)
//      book: O(N)
// Space: O(N)
class MyCalendarThree {
    map<int, int> m;
public:
    MyCalendarThree() {}
    int book(int start, int end) {
        m[start]++;
        m[end]--;
        int ans = 1, cur = 0;
        for (auto &[p, d] : m) {
            ans = max(ans, cur += d);
        }
        return ans;
    }
};