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.


 

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.
 

Note:

  1. You may assume the interval's end point is always bigger than its start point.
  2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

Solution:

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        int n = intervals.length;
        if (n == 0) return 0;
        Arrays.sort(intervals, (a, b) -> (a[0] - b[0] == 0? a[1] - b[1] : a[0] - b[0]));
        int longest = 1;
        int[] dp = new int[n];
        dp[0] = 1;
        for (int j = 1; j < n; j ++) {
            for (int i = 0; i < j; i ++) {
                if (intervals[j][0] >= intervals[i][1]) {
                    dp[j] = Math.max(dp[j], 1 + dp[i]);
                }
            }
            longest = Math.max(dp[j], longest);
        }
        return n - longest;
    }
}