"""
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.
"""
from typing import List


class Solution:
	def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
		if len(intervals) < 2:
			return 0
		intervals.sort()

		count, last_included = 0, 0
		for i in range(1, len(intervals)):
			if intervals[i][0] < intervals[last_included][1]:
				count += 1
				if intervals[i][1] < intervals[last_included][1]:
					last_included = i

			else:
				last_included = i

		return count


if __name__ == '__main__':
	S = Solution()

	print('-> 1: ', S.eraseOverlapIntervals([[1, 2], [2, 3], [3, 4], [1, 3]]))
	print('-> 2: ', S.eraseOverlapIntervals([[1, 2], [1, 2], [1, 2]]))
	print('-> 0: ', S.eraseOverlapIntervals([[1, 2], [2, 3]]))
	print('-> 2: ', S.eraseOverlapIntervals([[1, 100], [11, 22], [1, 11], [2, 12]]))
