Insert Interval

Problem Statement

You are given an array of non-overlapping intervals intervals where intervals[i] = [start_i, end_i] represents the start and the end time of the ith interval. intervals is initially sorted in ascending order by start_i.

You are given another interval newInterval = [start, end].

Insert newInterval into intervals such that intervals is still sorted in ascending order by start_i and also intervals still does not have any overlapping intervals. You may merge the overlapping intervals if needed.

Return intervals after adding newInterval.

Note: Intervals are non-overlapping if they have no common point. For example, [1,2] and [3,4] are non-overlapping, but [1,2] and [2,3] are overlapping.

Example 1:

Input: intervals = [[1,3],[4,6]], newInterval = [2,5]


Output: [[1,6]]

Example 2:

Input: intervals = [[1,2],[3,5],[9,10]], newInterval = [6,7]


Output: [[1,2],[3,5],[6,7],[9,10]]

Constraints:

  • 0 <= intervals.length <= 1000
  • newInterval.length == intervals[i].length == 2
  • 0 <= start <= end <= 1000

You should aim for a solution with O(n) time and O(1) space.

Hint 1

Iterate through the intervals and compare the new interval with each existing interval.

Hint 2

Consider three cases: the new interval is completely before, completely after, or overlapping with the current interval.

Hint 3

Merge overlapping intervals as needed.

Solution

def insert(intervals: list[list[int]], new_interval: list[int]) -> list[list[int]]:
    n = len(intervals)
    i = 0
    res = []


    while i < n and intervals[i][1] < new_interval[0]:
        res.append(intervals[i])
        i += 1


    while i < n and new_interval[1] >= intervals[i][0]:
        new_interval[0] = min(new_interval[0], intervals[i][0])
        new_interval[1] = max(new_interval[1], intervals[i][1])
        i += 1
    res.append(new_interval)


    while i < n:
        res.append(intervals[i])
        i += 1


    return res

Time complexity: O(n) Space complexity: O(n)