056. Merge Intervals

At a Glance

• Topic: Array
• Pattern: Analyze Pattern
• Difficulty: Medium
• LeetCode: 056

Problem Statement

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

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].

Example 2:

Input: intervals = [[1,4],[4,5]] Output: [[1,5]] Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Example 3:

Input: intervals = [[4,7],[1,4]] Output: [[1,7]] Explanation: Intervals [1,4] and [4,7] are considered overlapping.

Constraints:

1 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 104

Approach & Solution Steps

• Compare all pairs — O(n²).
• Sort + sweep — O(n log n) time / O(n) output space. <- pick this in interview.

Optimal JS Solution

function merge(intervals) {
	if (intervals.length === 0) return intervals;
	const sorted = [...intervals].sort((left, right) => left[0] - right[0]);
	const merged = [sorted[0]];
	for (let index = 1; index < sorted.length; index++) {
		const lastIndex = merged.length - 1;
		const [currentStart, currentEnd] = sorted[index];
		if (currentStart <= merged[lastIndex][1]) {
			merged[lastIndex][1] = Math.max(merged[lastIndex][1], currentEnd);
		} else {
			merged.push([currentStart, currentEnd]);
		}
	}
	return merged;
}

Edge Cases & Pitfalls

• Always consider empty or null inputs.
• Watch out for off-by-one index errors.