Merge Intervals in Python and TypeScript, Explained

The Problem

You get a list of intervals, each a pair [start, end]. Some of them overlap. Return a list where every group of overlapping intervals has been combined into one interval that covers the whole group, and no two intervals in the result overlap.

Example 1

Input: [[9, 12], [1, 4], [3, 6], [8, 9]]. Output: [[1, 6], [8, 12]]. [1, 4] and [3, 6] overlap, so they become [1, 6]. [8, 9] and [9, 12] touch at 9, which this version counts as overlapping.

Example 2

Input: [[1, 10], [2, 3], [4, 5]]. Output: [[1, 10]]. The first interval swallows the other two completely. This is why you take the maximum of the two end values instead of just copying the new end.

Assumptions to confirm

• Do intervals that only touch, like [1, 2] and [2, 3], count as overlapping? Ask, because it changes <= to <.
• Is the input already sorted? Usually not.
• May you change the input list, or should you return a new one? The solutions below leave the input untouched.

Brute Force First

A naive approach compares every pair of intervals and merges any two that overlap, then repeats until nothing changes. Each pass is O(n²), and you may need several passes, so it is slow and fiddly to get right. Mention it briefly, then point out the key insight: once intervals are sorted by start, any interval can only overlap the one you merged just before it.

The Optimal Approach: Sort, Then Sweep

Python and TypeScript Solutions

Python

def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:
    if not intervals:
        return []
    intervals = sorted(intervals, key=lambda iv: iv[0])
    merged = [intervals[0][:]]
    for start, end in intervals[1:]:
        last = merged[-1]
        if start <= last[1]:          # overlaps (or touches) the last block
            last[1] = max(last[1], end)
        else:
            merged.append([start, end])
    return merged

TypeScript

function mergeIntervals(intervals: number[][]): number[][] {
  if (intervals.length === 0) return [];
  const sorted = [...intervals].sort((a, b) => a[0] - b[0]);
  const merged: number[][] = [[...sorted[0]]];
  for (const [start, end] of sorted.slice(1)) {
    const last = merged[merged.length - 1];
    if (start <= last[1]) {
      last[1] = Math.max(last[1], end); // overlaps (or touches) the last block
    } else {
      merged.push([start, end]);
    }
  }
  return merged;
}

In TypeScript, pass a comparator to sort. Without one, JavaScript sorts arrays as strings, so [10, 12] would come before [9, 11]. That bug is a favourite follow-up question.

Complexity Compared

How to Talk Through It in an Interview

1. Clarify: "Do touching intervals like [1, 2] and [2, 3] merge? Is the input sorted? Can I return a new list?"
2. Brute force: "I could compare every pair and merge until nothing changes, but that is at least O(n²)."
3. Optimise: "If I sort by start, overlapping intervals become neighbours, so one pass is enough. The sort dominates at O(n log n)."
4. Code: sort, seed the result, loop with one comparison. Say out loud why you use the maximum of the ends.
5. Test: run the nested case [[1, 10], [2, 3]], a single interval, an empty list and two intervals with a gap.

Edge Cases and Follow-Ups

Practise It in a Mock Coding Round

Reading a solution is not the same as producing it with someone watching. In a real round you have to clarify the problem, say the brute force out loud, and keep talking while you type. MockIF runs coding rounds with a voice AI interviewer in a real in-browser editor, in Python and TypeScript. It runs your code against tests, asks you to explain your approach, follows up on edge cases and complexity, then scores the round and gives you specific feedback. A session costs 2 credits. It is free to start with 5 credits, then monthly or 3-monthly plans, or one-off credit packs (see pricing). Read more about how a mock coding interview works.

Frequently Asked Questions