253_MeetingRoomsII253.java

/**
 * Given an array of meeting time intervals consisting of start and end times
 * [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms
 * required.
 *
 * For example,
 * Given [[0, 30],[5, 10],[15, 20]],
 * return 2.
 *
 */

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */

public class MeetingRoomsII253 {
    public int minMeetingRooms(Interval[] intervals) {
        if (intervals == null || intervals.length == 0) return 0;

        Comparator<Interval> comparator = new Comparator<Interval>() {
            @Override
            public int compare(Interval i1, Interval i2) {
                return Integer.compare(i1.start, i2.start);
            }
        };

        Arrays.sort(intervals, comparator);

        Set<List<Interval>> set = new HashSet<>();

        for (Interval i: intervals) {
            boolean createNewList = true;
            for (List<Interval> l: set) {
                Interval ii = l.get(l.size()-1);
                if (ii.end <= i.start) {
                    l.add(i);
                    createNewList = false;
                    break;
                }
            }

            if (createNewList) {
                List<Interval> newList = new ArrayList<>();
                newList.add(i);
                set.add(newList);
            }
        }

        return set.size();
    }


    /**
     * https://leetcode.com/problems/meeting-rooms-ii/discuss/67855/Explanation-of-%22Super-Easy-Java-Solution-Beats-98.8%22-from-@pinkfloyda
     */
    public int minMeetingRooms2(Interval[] intervals) {
        int[] starts = new int[intervals.length];
        int[] ends = new int[intervals.length];
        for(int i=0; i<intervals.length; i++) {
            starts[i] = intervals[i].start;
            ends[i] = intervals[i].end;
        }
        Arrays.sort(starts);
        Arrays.sort(ends);
        int rooms = 0;
        int endsItr = 0;
        for(int i=0; i<starts.length; i++) {
            if(starts[i]<ends[endsItr])
                rooms++;
            else
                endsItr++;
        }
        return rooms;
    }


    /**
     * https://leetcode.com/problems/meeting-rooms-ii/discuss/67857/AC-Java-solution-using-min-heap
     */
    public int minMeetingRooms3(Interval[] intervals) {
        if (intervals == null || intervals.length == 0)
            return 0;

        // Sort the intervals by start time
        Arrays.sort(intervals, new Comparator<Interval>() {
            public int compare(Interval a, Interval b) { return a.start - b.start; }
        });

        // Use a min heap to track the minimum end time of merged intervals
        PriorityQueue<Interval> heap = new PriorityQueue<Interval>(intervals.length, new Comparator<Interval>() {
            public int compare(Interval a, Interval b) { return a.end - b.end; }
        });

        // start with the first meeting, put it to a meeting room
        heap.offer(intervals[0]);

        for (int i = 1; i < intervals.length; i++) {
            // get the meeting room that finishes earliest
            Interval interval = heap.poll();

            if (intervals[i].start >= interval.end) {
                // if the current meeting starts right after
                // there's no need for a new room, merge the interval
                interval.end = intervals[i].end;
            } else {
                // otherwise, this meeting needs a new room
                heap.offer(intervals[i]);
            }

            // don't forget to put the meeting room back
            heap.offer(interval);
        }

        return heap.size();
    }

}