Leetcode: Non-overlapping Intervals

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.

Actually, the problem is the same as "Given a collection of intervals, find the maximum number of intervals that are non-overlapping." (the classic Greedy problem: Interval Scheduling). With the solution to that problem, guess how do we get the minimum number of intervals to remove? : )

Sorting Interval.end in ascending order is O(nlogn), then traverse intervals array to get the maximum number of non-overlapping intervals is O(n). Total is O(nlogn).

开始的时候想岔了，以为是要求同一时刻overlap的最多interval数，但仔细想一想就发现不对，应该是non-overlap的interval的最大数目

1. Best solution: sorted by interval end

case 1 add current interval as another non-overlapping interval, case 2 and case 3 all get rid of the current interval

/**
 * 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 Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        if (intervals.length == 0) return 0;
        int nonOverlap = 1;
        int seq = 0;
        Arrays.sort(intervals, new Comparator<Interval>() {
            public int compare(Interval i1, Interval i2) {
                return i1.end - i2.end;
            }
        });
        for (int i=1; i<intervals.length; i++) {
            if (intervals[i].start >= intervals[seq].end) {
                seq = i;
                nonOverlap++;
            }
        }
        return intervals.length - nonOverlap;
    }
}

 Comparator can also be rewritten as

Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[1], i2[1]));

2. Alternatives(not the best): sort by interval start

case 1 add current interval as another non-overlapping interval, case 2 update the previous non-overlapping interval with the current one, and case 3 get rid of the current interval. So more cases need to be processed than sorted by interval end

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if (intervals.length < 1) return 0;
        int seq = 0;
        int nonOverlap = 1;
        
        Arrays.sort(intervals, (i1, i2) -> Integer.compare(i1[0], i2[0]));
        
        for (int i = 0; i < intervals.length; i ++) {
            if (intervals[i][0] >= intervals[seq][1]) {
                seq = i;
                nonOverlap ++;
            }
            else if (intervals[i][1] <= intervals[seq][1]) {
                seq = i;
            }
        }
        
        return intervals.length - nonOverlap;
    }
}