public int minMeetingRooms1(Interval[] intervals) {
        int[] start = new int[intervals.length];
        int[] end = new int[intervals.length];

        for (int i = 0; i < intervals.length; i++) {
            start[i] = intervals[i].start;
            end[i] = intervals[i].end;
        }

        Arrays.sort(start);
        Arrays.sort(end);

        int j = 0;
        int rooms = 0;
        for (int i = 0; i < start.length; i++) {
            if (start[i] < end[j]) {
                rooms++;
            } else {
                j++;
            }
        }
        return rooms;
    }
Date 05/01/2016
@author Tushar Roy

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.

Both methods have time comlexity of nlogn
Method 1 has space complexity of O(1)

https://leetcode.com/problems/meeting-rooms-ii/
src/com/interview/array/MeetingRooms.java