"""
435. Non-overlapping Intervals
Difficulty: Medium
https://leetcode.com/problems/non-overlapping-intervals/

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Given an array of intervals intervals where intervals[i] = [starti,
endi], return the minimum number of intervals you need to remove to
make the rest of the intervals non-overlapping.

Note that intervals which only touch at a point are non-overlapping.
For example, [1, 2] and [2, 3] are non-overlapping.



Example 1:

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are
non-overlapping.

Example 2:

Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the
intervals non-overlapping.

Example 3:

Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since
they're already non-overlapping.



Constraints:

        • 1 <= intervals.length <= 10^5

        • intervals[i].length == 2

        • -5 * 10^4 <= starti < endi <= 5 * 10^4
"""


class Solution:
    def eraseOverlapIntervals(self, intervals: list[list[int]]) -> int:
        intervals.sort(key=lambda x: x[0])
        result = []

        for start, end in intervals:
            # Use '<' to avoid intervals which touch
            if result and start < result[-1][1]:
                result[-1][1] = min(result[-1][1], end)
            else:
                result.append([start, end])

        # Difference between original and result is to remove
        return len(intervals) - len(result)