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.
 *
 * <p>For example, Given [[0, 30],[5, 10],[15, 20]], return false.
 *
 * <p>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;
  }
}