Skip to main content

57. Insert Interval

mediumAsked at Airbnb

Splice a new booking into a sorted availability list without disturbing existing reservations — Airbnb's calendar service does this on every Instant Book confirmation that lands between two blocked windows.

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

Problem

You are given an array of non-overlapping intervals sorted by start time and a new interval. Insert the new interval and merge any overlapping ones, returning the resulting sorted array of non-overlapping intervals.

Constraints

  • 0 <= intervals.length <= 10^4
  • intervals[i].length == 2
  • 0 <= start_i <= end_i <= 10^5
  • intervals is sorted in ascending order by start_i
  • newInterval.length == 2

Examples

Example 1

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

Explanation: The new interval [2,5] overlaps [1,3] and merges into [1,5].

Example 2

Input
intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output
[[1,2],[3,10],[12,16]]

Explanation: [4,8] overlaps [3,5], [6,7], and [8,10], merging them into [3,10].

Approaches

1. Brute force (append and re-merge)

Append the new interval, then run the full Merge Intervals sort-and-sweep. Correct but wastes the fact that the input is already sorted.

Time
O(n log n)
Space
O(n)
function insert(intervals, newInterval) {
  const all = [...intervals, newInterval].sort((a, b) => a[0] - b[0]);
  const merged = [all[0]];
  for (let i = 1; i < all.length; i++) {
    const last = merged[merged.length - 1];
    if (all[i][0] <= last[1]) {
      last[1] = Math.max(last[1], all[i][1]);
    } else {
      merged.push(all[i]);
    }
  }
  return merged;
}

Tradeoff:

2. Single-pass linear insert

Three-phase sweep: copy all intervals that end before newInterval starts, merge all that overlap, then copy the rest. Exploits sorted order for O(n) time.

Time
O(n)
Space
O(n)
function insert(intervals, newInterval) {
  const result = [];
  let i = 0;
  const n = intervals.length;
  while (i < n && intervals[i][1] < newInterval[0]) {
    result.push(intervals[i++]);
  }
  while (i < n && intervals[i][0] <= newInterval[1]) {
    newInterval[0] = Math.min(newInterval[0], intervals[i][0]);
    newInterval[1] = Math.max(newInterval[1], intervals[i][1]);
    i++;
  }
  result.push(newInterval);
  while (i < n) result.push(intervals[i++]);
  return result;
}

Tradeoff:

Airbnb-specific tips

Interviewers at Airbnb often give this right after Merge Intervals to see if you can exploit the sorted guarantee. Explicitly state the three phases before coding — they want to hear that you recognize no sorting is needed. Common trap: off-by-one on the overlap boundary conditions (use strict < for ends-before and <= for starts-after-merge).

Solve it now

Free. No sign-up. Python and JavaScript run instantly in your browser.

Output

Press Run or Cmd+Enter to execute

Practice these live with InterviewChamp.AI

Drill Insert Interval and other Airbnb interview questions under real-loop conditions with instant feedback on your reasoning, complexity claims, and code.

Practice these live with InterviewChamp.AI →