[Leetcode] 0435. Non-overlapping Intervals

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内容目录

Medium

Given an array of intervals intervals where intervals[i] = [start _i , end _i ], 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 <= start _i < end _i <= 5 * 10 ⁴

Python

				
					# time complexity: O(nlogn) # space complexity: O(n) from typing import List class Solution: def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int: intervals.sort(key=lambda x : x[1]) k = -float("inf") ans = 0 for x, y in intervals: if x >= k: k = y else: ans += 1 return ans intervals = [[1, 2] , [2, 3], [3, 4], [1, 3]] print(Solution().eraseOverlapIntervals(intervals))

Medium

Python

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