```
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:

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.

```

class Solution(object):
    def eraseOverlapIntervals(self, intervals):
        if len(intervals) == 1 or len(intervals) == 0:
            return 0
        interval = sorted(intervals, key=lambda x: x[1])
        curr = interval[0]
        ans = 1
        for val in interval:
            if curr[1]<=val[0]:
                curr = val
                ans += 1
        return len(interval) - ans
        
