22. Merge Intervals

easyAsked at Doordash

Collapse overlapping time windows into the fewest spans — Doordash uses this interval pattern directly in delivery time-window consolidation and Dasher shift-block scheduling.

By Alex Chen, Founder, InterviewChamp.AI · Last verified 2026-05-26

Problem

Given an array of intervals where intervals[i] = [start_i, end_i], merge all overlapping intervals and return an array of the non-overlapping intervals that cover all the intervals in the input.

Constraints

1 <= intervals.length <= 10^4
intervals[i].length == 2
0 <= start_i <= end_i <= 10^4

Examples

Example 1

Input

intervals = [[1,3],[2,6],[8,10],[15,18]]

Output

[[1,6],[8,10],[15,18]]

Explanation: Intervals [1,3] and [2,6] overlap; merged to [1,6].

Example 2

Input

intervals = [[1,4],[4,5]]

Output

[[1,5]]

Explanation: Touching at 4 counts as overlapping.

Approaches

1. Brute force nested comparison

Compare every pair of intervals to find overlaps; merge and restart. O(n^2) passes until stable.

Time: O(n^2)
Space: O(n)

function merge(intervals) {
  let changed = true;
  while (changed) {
    changed = false;
    const result = [];
    const used = new Array(intervals.length).fill(false);
    for (let i = 0; i < intervals.length; i++) {
      if (used[i]) continue;
      let [s, e] = intervals[i];
      for (let j = i + 1; j < intervals.length; j++) {
        if (used[j]) continue;
        const [s2, e2] = intervals[j];
        if (s2 <= e && e2 >= s) {
          s = Math.min(s, s2); e = Math.max(e, e2);
          used[j] = true; changed = true;
        }
      }
      result.push([s, e]);
    }
    intervals = result;
  }
  return intervals;
}

Tradeoff:

2. Sort then linear scan

Sort by start time; single pass merging the current interval into the last merged interval whenever they overlap. Classic greedy approach.

Time: O(n log n)
Space: O(n)

function merge(intervals) {
  intervals.sort((a, b) => a[0] - b[0]);
  const merged = [intervals[0]];
  for (let i = 1; i < intervals.length; i++) {
    const last = merged[merged.length - 1];
    if (intervals[i][0] <= last[1]) {
      last[1] = Math.max(last[1], intervals[i][1]);
    } else {
      merged.push(intervals[i]);
    }
  }
  return merged;
}

Tradeoff:

Doordash-specific tips

Doordash interviewers expect you to immediately connect intervals to real scheduling — 'how would you handle Dasher availability windows that span midnight?' The follow-up is almost always: what if intervals arrive as a stream? That pushes you toward a sorted structure (BST or sorted list with binary search insertion). Name the pattern — greedy after sort — before diving into code.

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