Problem Overview & Intuition

By sorting intervals by their start coordinate, any overlapping intervals will appear consecutively. Comparing the current interval start with the previous interval end resolves all merges in a single linear pass.

Step-by-Step Approach

1. Sort intervals by starting value a[0] - b[0].
2. Initialize result = [intervals[0]].
3. Iterate from i = 1 to intervals.length - 1.
4. If intervals[i][0] <= last[1]: merge by last[1] = Math.max(last[1], intervals[i][1]).
5. Otherwise, push intervals[i] to result.
6. Return result.

Optimal Implementation (JavaScript)

function mergeIntervals(intervals) {
  if (!intervals.length) return [];
  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]) {
      last[1] = Math.max(last[1], current[1]);
    } else {
      result.push(current);
    }
  }
  return result;
}

Complexity Analysis