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 <= 105
intervals[i].length == 2
-5 * 104 <= starti < endi <= 5 * 104
Solution :
class Solution: def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int: intervals.sort() ans = 0 end_prev = intervals[0][1] for s,e in intervals[1:]: if s >= end_prev : end_prev = e else: ans += 1 end_prev = min(e, end_prev) return ans
Explaination :
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