## DEV Community

Abhishek Chaudhary

Posted on

# Merge Intervals

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
``````