DSA TrainerInsert Interval
Given a sorted array of non-overlapping intervals and a new interval, insert it into the correct position and merge any overlapping intervals. Return the result.
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Problem
You are given an array of non-overlapping intervals sorted in ascending order. You are also given an interval newInterval. Insert newInterval into intervals such that intervals is still sorted in ascending order and has no overlapping intervals. Return intervals after the insertion.
Input
A sorted array of non-overlapping intervals `intervals` (sorted by start time) and a `newInterval` to insert.
Output
The intervals array after inserting `newInterval` and merging all overlapping intervals.
Examples
Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]
[2,5] overlaps [1,3] (2 ≤ 3). They merge into [1,5]. [6,9] doesn't overlap.
Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
[4,8] overlaps [3,5], [6,7], and [8,10]. They all merge into [3,10].
The brute-force approach
Insert newInterval into the array, sort everything by start time, then run the standard merge-intervals algorithm.
intervals.append(newInterval)
intervals.sort(key=lambda x: x[0])
# now run merge intervalsO(n log n) for the sort — but the input is already sorted. You can solve this in O(n) with a single pass since the existing intervals are guaranteed non-overlapping and sorted.
Spotting the pattern
This is a Intervals problem. The key question to ask yourself:
When does an existing interval come entirely before newInterval? Entirely after? When do they overlap?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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