NonOverlappingIntervals435.java

/**
 * Given a collection of intervals, find the minimum number of intervals you
 * need to remove to make the rest of the intervals non-overlapping.
 * 
 * Note:
 * You may assume the interval's end point is always bigger than its start point.
 * Intervals like [1,2] and [2,3] have borders "touching" but they don't
 * overlap each other.
 * 
 * Example 1:
 * Input: [ [1,2], [2,3], [3,4], [1,3] ]
 * Output: 1
 * Explanation: [1,3] can be removed and the rest of intervals are
 * non-overlapping.
 * 
 * Example 2:
 * Input: [ [1,2], [1,2], [1,2] ]
 * Output: 2
 * Explanation: You need to remove two [1,2] to make the rest of intervals
 * non-overlapping.
 * 
 * Example 3:
 * Input: [ [1,2], [2,3] ]
 * Output: 0
 * Explanation: You don't need to remove any of the intervals since they're
 * already non-overlapping.
 */

public class NonOverlappingIntervals435 {

    private static Comparator<Interval> comp = new Comparator<Interval>() {
        @Override
        public int compare(Interval i1, Interval i2) {
            int diff = Integer.compare(i1.start, i2.start);
            if (diff != 0) return diff;
            return Integer.compare(i2.end-i2.start, i1.end-i1.start);
        }
    };

    public int eraseOverlapIntervals(Interval[] intervals) {
        if (intervals == null || intervals.length == 0) return 0;
        Arrays.sort(intervals, comp);
        Interval inv = intervals[0];
        int res = 0;
        for (int i=1; i<intervals.length; i++) {
            Interval curr = intervals[i];
            if (curr.start == inv.start) {
                inv = curr;
                res++;
            } else if (curr.start < inv.end) {
                if (curr.end < inv.end) {
                    inv = curr;
                }
                res++;
            } else {
                inv = curr;
            }
        }
        return res;
    }

}