interval-optimizer.js

function solution(input) {
  if (input.length === 0) {
    return [];
  }

  // sort by first element of each interval
  // O(n log n)
  intervals = input.slice(0);
  intervals.sort(([a], [b]) => a - b);

  // sweep through intervals and chain them together
  // if chain broken, flush into results and continue
  // O(n)
  var result = [];
  var [start, end] = intervals.shift();
  intervals.forEach(([s, e]) => {
    if (s <= end) {
      end = Math.max(e, end);
    } else {
      result.push([start, end]);
      [start, end] = [s, e];
    }
  });

  // the last chain would never be broken, so flush it!
  result.push([start, end]);

  return result;
}

var tests = [
  [[2, 5], [1, 10], [11, 15], [1, 3]],
  [],
  [[4, 5], [1, 2], [2, 3]],
  [[1, 10], [2, 2]]
];

tests.forEach(t => console.log(t, '=>', solution(t)));