Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.
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.
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/non-overlapping-intervals # Definition for an interval. # class Interval(object): # def __init__(self, s=0, e=0): # self.start = s # self.end = e class Solution(object): def eraseOverlapIntervals(self, intervals): """ :type intervals: List[Interval] :rtype: int """