What is the brute force approach and complexity for Insert Interval?

Add the new interval, sort everything, merge like Merge Intervals with time complexity O(N log N) and space complexity O(N).

What is the optimal approach and complexity for Insert Interval?

Copy the left side, swallow the overlaps into the new interval, copy the right side with optimal time complexity O(N) and space complexity O(N).

How do you identify the pattern for Insert Interval?

Look for key signals in the prompt: "already sorted by start" indicates The sort step of Merge Intervals is already done for us — don't pay for it again.; "no two intervals overlap" indicates Only the new interval can cause a merge, so overlaps form one run around it.; "insert and merge if needed" indicates It is the Merge Intervals step, but with only one interval that can grow..

Insert Interval

Medium·LeetCode #57 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

You get a list of intervals [start, end] that is already sorted by start, and no two of them overlap. You also get one more interval, newInterval.

Put newInterval into the list. If it overlaps some intervals, join them all into one. Return the new list, still sorted and still with no overlaps.

Intervals that only touch (like [1,5] and [5,7]) count as overlapping.

Input

A sorted, non-overlapping list intervals, and one interval newInterval = [start, end].

Output

The list after inserting and merging, sorted by start.

Example 1

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

Output: [[1,5],[6,9]]

[2,5] overlaps [1,3], so they become [1,5]. [6,9] starts after 5, so it stays.

Example 2

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] (it touches 8). Together they cover [3,10].

Constraints

• 0 ≤ intervals.length ≤ 10^4
• intervals[i].length == 2, and 0 ≤ start ≤ end ≤ 10^5
• intervals is sorted by start and has no overlaps
• newInterval.length == 2, and 0 ≤ start ≤ end ≤ 10^5

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“already sorted by start”

→ The sort step of Merge Intervals is already done for us — don't pay for it again.

“no two intervals overlap”

→ Only the new interval can cause a merge, so overlaps form one run around it.

“insert and merge if needed”

→ It is the Merge Intervals step, but with only one interval that can grow.

Approach Complexity Ladder

Brute Force — append and run Merge IntervalsO(N log N)O(N)

Add the new interval, sort everything, merge like Merge Intervals

Optimal — three phases in one passO(N)O(N)

Copy the left side, swallow the overlaps into the new interval, copy the right side

Brute Force Intuition

Push newInterval onto a copy of the list, sort by start, then merge exactly as in Merge Intervals: compare each interval with the last merged one and stretch its end.

Recognition rule for this sub-pattern:

Sort by start. If next.start ≤ current.end, they overlap — extend current.end. Otherwise start a new interval.

Remember this · Intervals

Trigger

Input is [start, end] pairs; overlap, merge, insert, meeting rooms, free time, minimum removals.

Think

After sorting, do I only ever need to compare against the most recent interval?

Key question

Sort by start or by end — which one makes the greedy choice obvious?

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