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64 lines
1.7 KiB
Python
64 lines
1.7 KiB
Python
"""
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435. Non-overlapping Intervals
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Difficulty: Medium
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https://leetcode.com/problems/non-overlapping-intervals/
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──────────────────────────────────────────────────
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Given an array of intervals intervals where intervals[i] = [starti,
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endi], return the minimum number of intervals you need to remove to
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make the rest of the intervals non-overlapping.
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Note that intervals which only touch at a point are non-overlapping.
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For example, [1, 2] and [2, 3] are non-overlapping.
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Example 1:
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Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
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Output: 1
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Explanation: [1,3] can be removed and the rest of the intervals are
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non-overlapping.
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Example 2:
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Input: intervals = [[1,2],[1,2],[1,2]]
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Output: 2
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Explanation: You need to remove two [1,2] to make the rest of the
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intervals non-overlapping.
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Example 3:
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Input: intervals = [[1,2],[2,3]]
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Output: 0
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Explanation: You don't need to remove any of the intervals since
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they're already non-overlapping.
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Constraints:
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• 1 <= intervals.length <= 10^5
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• intervals[i].length == 2
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• -5 * 10^4 <= starti < endi <= 5 * 10^4
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"""
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class Solution:
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def eraseOverlapIntervals(self, intervals: list[list[int]]) -> int:
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intervals.sort(key=lambda x: x[0])
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result = []
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for start, end in intervals:
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# Use '<' to avoid intervals which touch
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if result and start < result[-1][1]:
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result[-1][1] = min(result[-1][1], end)
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else:
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result.append([start, end])
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# Difference between original and result is to remove
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return len(intervals) - len(result)
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