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56. Merge Intervals

mediumAsked at Airbnb

Collapse overlapping date ranges into clean availability windows — the exact interval-merge logic Airbnb's calendar engine runs every time a host blocks dates or a guest extends a stay.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

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

Example 2

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

Explanation: Intervals [1,4] and [4,5] are considered overlapping (touching boundary).

Approaches

1. Brute force

For each interval, scan all others to find overlaps and merge. Repeat 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 sweep

Sort by start time, then do a single left-to-right sweep: extend the current interval's end or start a new one. Classic greedy interval merge.

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];
    const [start, end] = intervals[i];
    if (start <= last[1]) {
      last[1] = Math.max(last[1], end);
    } else {
      merged.push([start, end]);
    }
  }
  return merged;
}

Tradeoff:

Airbnb-specific tips

Airbnb recruiters confirm this appears in nearly every on-site loop — framed as 'consolidate blocked-off dates on a host calendar.' They grade on edge cases: adjacent intervals (touching but not overlapping), a single-element input, and already-sorted vs unsorted input. Talk through the sort step first and explain why it reduces the problem to a single pass.

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