17. Merge Intervals

mediumAsked at Glassdoor

Glassdoor's salary-range bands overlap across job titles and geographies — merging those intervals cleanly is a real backend task, which is why this sort-and-sweep problem shows up regularly in their coding screens.

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

Problem

Given an array of intervals where intervals[i] = [starti, endi], 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 <= starti <= endi <= 10^4

Examples

Example 1

Input

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

Output

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

Explanation: [1,3] and [2,6] overlap because 2 <= 3, so they merge into [1,6].

Example 2

Input

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

Output

[[1,5]]

Explanation: Intervals that touch at a single point are considered overlapping.

Approaches

1. Brute force — repeated linear scan

Repeatedly scan the list and merge any overlapping pair until no merges occur. O(n^2) passes, impractical for large inputs.

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

function merge(intervals) {
  let changed = true;
  while (changed) {
    changed = false;
    const next = [];
    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;
        }
      }
      next.push([s, e]);
    }
    intervals = next;
  }
  return intervals;
}

Tradeoff:

2. Sort then sweep

Sort intervals by start time. Walk once: if the current interval overlaps the last merged one, extend it; otherwise push a new entry. O(n log n) sort dominates.

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

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

Tradeoff:

Glassdoor-specific tips

Glassdoor tests merge-intervals specifically because salary data arrives in overlapping bands from multiple sources and must be deduped before display. Interviewers expect you to sort first — if you skip that step they'll ask why. Mention the edge case where two intervals touch exactly at one endpoint (e.g., [1,4] and [4,5]) and confirm your overlap condition handles it with <=, not <.

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