435. Non-overlapping Intervals

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

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

Solution

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)