729. My Calendar I

Description

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

Your class will have the 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 double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a double 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)

Constraints

• 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].

Approach

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Examples

Example 1

Input:

["MyCalendar","book","book","book"]

[[],[10,20],[15,25],[20,30]]

Output:

[null,true,false,true]

Explanation:

MyCalendar();

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

MyCalendar.book(15, 25); // returns false

MyCalendar.book(20, 30); // returns true

The first event can be booked. The second can't because time 15 is already booked by another event.

The third event can be booked, as the first event takes every time less than 20, but not including 20.

Solutions

Solution 1

/**
 * Time complexity : O(N^2), where N is the number of events booked. For each 
 *    new event, we process every previous event to decide whether the new 
 *    event can be booked. This leads to ∑(k,N) O(k) = O(N^2) complexity.
 * Space complexity : O(N), the size of the calendar.
 */
 
 public class MyCalendar {
    List<int[]> calendar;

    MyCalendar() {
        calendar = new ArrayList();
    }

    public boolean book(int start, int end) {
        for (int[] iv: calendar) {
            if (iv[0] < end && start < iv[1]) return false;
        }
        calendar.add(new int[]{start, end});
        return true;
    }
}

Solution 2

/**
 * Time complexity : O(NlogN), where N is the number of events booked. 
 *    For each new event, we search that the event is legal in 
 *    O(logN) time, then insert it in O(1) time.
 * Space complexity : O(N), the size of the data structures used.
 */
 
 class MyCalendar {
    TreeMap<Integer, Integer> calendar;

    MyCalendar() {
        calendar = new TreeMap();
    }

    public boolean book(int start, int end) {
        Integer prev = calendar.floorKey(start),
                next = calendar.ceilingKey(start);
        if ((prev == null || calendar.get(prev) <= start) &&
                (next == null || end <= next)) {
            calendar.put(start, end);
            return true;
        }
        return false;
    }
}

Solution 3

/**
 * Time complexity : 
 * Space complexity : 
 */

class ListNode {
    int start, end;
    ListNode next;
    
    ListNode(int start, int end, ListNode next) {
        this.start = start;
        this.end = end;
        this.next = next;
    }
}

class MyCalendar {
    ListNode head;

    public MyCalendar() {
        ListNode tail = new ListNode(Integer.MAX_VALUE, Integer.MAX_VALUE, null);
        head = new ListNode(-1, -1, tail);
    }
    
    public boolean book(int start, int end) {
        ListNode curr = head, last = head;
        while(curr.start < start) {
            last = curr;
            curr = curr.next;
        }
        if(last.end > start || curr.start < end) {
            return false;
        }
        last.next = new ListNode(start, end, curr);
        return true;
    }
}

/**
 * Your MyCalendar object will be instantiated and called as such:
 * MyCalendar obj = new MyCalendar();
 * boolean param_1 = obj.book(start,end);
 */

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