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.

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: Dynamic programming

Sorting

Data structures

Greedy algorithms

Starter code: class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        

Solutions:
class Solution:

	def eraseOverlapIntervals(self, intervals):
		intervals.sort(key=lambda x: x.end)
		end = -sys.maxsize
		count = 0
		for i in intervals:
			if i.start < end:
				continue
			end = i.end
			count += 1
		return len(intervals) - count