MeetingRooms.java

package array;

import java.util.Arrays;

/**
 * Created by gouthamvidyapradhan on 27/11/2017.
 * Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all meetings.

 For example,
 Given [[0, 30],[5, 10],[15, 20]],
 return false.

 Solution: Sort the interval based on the start interval. Then, for every interval check if the previous interval
 ends before the start of the current interval.
 */
public class MeetingRooms {

    public static class Interval {
        int start;
        int end;
        Interval() { start = 0; end = 0; }
        Interval(int s, int e) { start = s; end = e; }
    }

    public static void main(String[] args) throws Exception{
        Interval i1 = new Interval(0, 30);
        Interval i2 = new Interval(5, 10);
        Interval i3 = new Interval(15, 20);
        Interval[] intervals = {i1, i2, i3};

        System.out.println(new MeetingRooms().canAttendMeetings(intervals));
    }

    public boolean canAttendMeetings(Interval[] intervals) {
        Arrays.sort(intervals, (a, b) -> Integer.compare(a.start, b.start));
        for(int i = 1; i < intervals.length; i ++){
            if(intervals[i].start < intervals[i - 1].end)
                return false;
        }
        return true;
    }

}