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22. Insert Interval

mediumAsked at Instacart

Insert a new interval into a sorted, non-overlapping list and merge as needed — Instacart uses this when a new delivery window is injected into an existing shopper schedule without disturbing confirmed slots.

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

Problem

You are given an array of non-overlapping intervals sorted in ascending order by start time, and a single newInterval to insert. Insert newInterval into intervals and merge if necessary. Return the resulting array of intervals, also sorted by start time with no overlapping intervals.

Constraints

  • 0 <= intervals.length <= 10^4
  • intervals[i].length == 2
  • 0 <= start_i <= end_i <= 10^5
  • intervals is sorted by start_i in ascending order
  • newInterval.length == 2
  • 0 <= newInterval[0] <= newInterval[1] <= 10^5

Examples

Example 1

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

Explanation: [2,5] overlaps [1,3], so they merge 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]]

Approaches

1. Insert then merge all

Append the new interval, sort by start, then run a standard merge-intervals pass.

Time
O(n log n)
Space
O(n)
function insert(intervals, newInterval) {
  intervals.push(newInterval);
  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];
    if (intervals[i][0] <= last[1]) {
      last[1] = Math.max(last[1], intervals[i][1]);
    } else {
      result.push(intervals[i]);
    }
  }
  return result;
}

Tradeoff:

2. Linear single pass

Since intervals is already sorted, walk through in three phases: copy all intervals that end before newInterval starts, merge all overlapping, copy all that start after newInterval ends.

Time
O(n)
Space
O(n)
function insert(intervals, newInterval) {
  const result = [];
  let i = 0;
  const n = intervals.length;

  // Phase 1: intervals completely before newInterval
  while (i < n && intervals[i][1] < newInterval[0]) {
    result.push(intervals[i++]);
  }

  // Phase 2: merge overlapping
  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);

  // Phase 3: intervals completely after newInterval
  while (i < n) result.push(intervals[i++]);

  return result;
}

Tradeoff:

Instacart-specific tips

Instacart schedules shoppers in pre-sorted time blocks. The three-phase linear pass is the expected answer — it exploits the sorted invariant and runs in O(n) versus O(n log n) for re-sorting. Interviewers specifically look for whether you handle the edge cases: new interval fully before all existing ones, fully after, or spanning all of them.

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