LeetCode [435] Non-overlapping Intervals

 435. Non-overlapping Intervals

Medium

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.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

class Solution { public int eraseOverlapIntervals(int[][] intervals) { if (intervals.length == 0) return 0; Arrays.sort(intervals, (a,b)->Integer.compare(a[1], b[1])); int end = intervals[0][1]; int count = 1; for (int i = 1; i < intervals.length; i++) { if (intervals[i][0] >= end) { end = intervals[i][1]; count++; } } return intervals.length - count; } }