## DEV Community

Abhishek Chaudhary

Posted on

# Insert Interval

You are given an array of non-overlapping intervals `intervals` where `intervals[i] = [starti, endi]` represent the start and the end of the `ith` interval and `intervals` is sorted in ascending order by `starti`. You are also given an interval `newInterval = [start, end]` that represents the start and end of another interval.

Insert `newInterval` into `intervals` such that `intervals` is still sorted in ascending order by `starti` and `intervals` still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return `intervals` after the insertion.

Example 1:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

Constraints:

• `0 <= intervals.length <= 104`
• `intervals[i].length == 2`
• `0 <= starti <= endi <= 105`
• `intervals` is sorted by `starti` in ascending order.
• `newInterval.length == 2`
• `0 <= start <= end <= 105`

SOLUTION:

``````import bisect

class Solution:
def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
bisect.insort(intervals, newInterval)
n = len(intervals)
i = 0
while i < len(intervals) - 1:
if intervals[i][1] >= intervals[i + 1][0]:
if intervals[i][1] <= intervals[i + 1][1]:
intervals[i][1] = intervals[i + 1][1]
intervals.pop(i + 1)
else:
i += 1
return intervals
``````