Merge Intervals - Netflix Interview Question & Solution

Merge Intervals

Netflix

Array

Sorting

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.

Example:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Solution Walkthrough

Overlaps are only easy to detect once the intervals are in a predictable order — so the first move is always to sort by start time. After that, a single linear pass can merge everything.

Sort intervals in ascending order by their start value.
Initialize the result with the first interval.
For each subsequent interval, compare it against the last interval already placed in the result:
- If the current interval's start is less than or equal to that last interval's end, they overlap — extend the last interval's end to the maximum of the two ends.
- Otherwise, there's a gap — push the current interval onto the result as a new, separate interval.

The sort dominates the cost at O(n log n); the merge pass itself is a single O(n) sweep. Overall time complexity is O(n log n), with O(n) space for the result (or O(log n) to O(n) depending on the sort's implementation).

function merge(intervals) {
    if (intervals.length <= 1) return intervals;

    // Sort by start time so overlaps are always adjacent
    intervals.sort((a, b) => a[0] - b[0]);

    const result = [intervals[0]];

    for (let i = 1; i < intervals.length; i++) {
        const current = intervals[i];
        const last = result[result.length - 1];

        if (current[0] <= last[1]) {
            // Overlaps with the last merged interval — extend it
            last[1] = Math.max(last[1], current[1]);
        } else {
            // No overlap — it's a new interval
            result.push(current);
        }
    }

    return result;
}

Related Algorithms

Merge Sort →