LC 435 - Non-overlapping Intervals

Question

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

Links

Question here and solution here

Solution

concept

1. We first sort the intervals (either by start or end, start will be a bit more intuitive)
2. we keep track the current ending point in prev_end
3. As we iterate through the intervals, we need to check:
   1. if the current interval is overlapping with the previous intervals
      1. by checking if start >= prev_end, if true means it is not overlapping. Since it is sorted, we are sure the current start is >= prev_start so this condition is enough.
      2. If true, we will update prev_end to the current interval, since they are not overlapping.
   2. if the current interval is overlapping:
      1. we need to remove the interval with larger end point to maximize the likelihood that the future intervals won’t be overlapping current intervals.

code

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        intervals.sort()
        prev_end = intervals[0][1]
        ans = 0

        for start, end in intervals[1:]:
            if start >= prev_end:
                prev_end = end
            else:
                ans += 1
                prev_end = min(prev_end, end)

        return ans

Complexity

time: $O(n \log n)$
space: $O(n)$