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435. Non Overlapping Intervals

Array Dynamic Programming Greedy Sorting

Problem - Non Overlapping Intervals

Medium

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 <= 105
  • intervals[i].length == 2
  • -5 * 104 <= starti < endi <= 5 * 104

Solutions

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class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort(key=lambda x: x[1])
        result = len(intervals)
        last_val = float('-inf')

        for start, end in intervals:
            if last_val <= start:
                result -= 1
                last_val = end

        return result

Submission Stats:

  • Runtime: 67 ms (84.11%)
  • Memory: 51.4 MB (22.31%)