Merge Intervals

#leetcode #dsa #theabbie

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints:

1 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 104

SOLUTION:

import bisect

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        points = []
        for l, r in intervals:
            bisect.insort(points, (l, 1))
            bisect.insort(points, (r + 1, -1))
        p = 0
        prev = None
        intervals = []
        for x, diff in points:
            p += diff
            if p == 0 and prev != None:
                if diff == 1:
                    intervals.append([prev, x])
                else:
                    intervals.append([prev, x - 1])
            if p == 1 and p - diff == 0:
                prev = x
        return intervals