How do you solve interval problems?

Intervals are pairs of start and end values. The first decision is the sort key. Merging overlapping intervals sorts by start. Maximizing non-overlapping intervals sorts by end. Using the wrong key leads to a correct-looking loop with the wrong answer.

function mergeIntervals(intervals) {
  if (intervals.length === 0) return [];
  const sorted = [...intervals].sort((a, b) => a[0] - b[0]);
  const result = [sorted[0]];

  for (let i = 1; i < sorted.length; i++) {
    const current = sorted[i];
    const last = result[result.length - 1];
    if (current[0] <= last[1]) {
      last[1] = Math.max(last[1], current[1]);
    } else {
      result.push(current);
    }
  }

  return result;
}

The main variants

• Merge overlapping intervals: sort by start and extend the last end
• Insert an interval: find the insertion point, then merge neighbours
• Meeting rooms: sort starts and ends separately, then sweep
• Maximum non-overlapping intervals: sort by end and take the earliest finishing
• Intersection of two interval lists: use two pointers

Decide whether touching intervals count as overlapping. The condition current[0] <= last[1] treats a shared endpoint as overlap; current[0] < last[1] treats it as separate. State the convention before coding.