17. Merge Intervals

mediumAsked at Wix

Collapse overlapping time ranges into non-overlapping segments — Wix applies this pattern to animation timeline collision detection and to calendar/booking widgets in the site builder.

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, so they merge 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

For each interval, repeatedly scan all others to find overlaps and merge. Repeat until no merges happen.

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 single-pass sweep

Sort by start time. Walk through once: if the current interval's start is within the last merged interval's end, extend the end; otherwise start a new merged interval.

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:

Wix-specific tips

Wix asks this in the context of scheduling and animation timelines. Mentioning 'sort to establish a global order, then a single greedy sweep' maps directly to how their animation engine resolves keyframe collisions — that framing earns extra points with Wix interviewers.

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