JavaScript ES5 version of the assignment

assignment.js

/**
 * finds the highest product of the three items passed in
 * @param arrayOfInts<type array>
 */
function highestProductOf3(arrayOfInts) {
  if (arrayOfInts.length < 3) {
    throw new Error('Less than 3 items!');
  }

  // we're going to start at the 3rd item (at index 2)
  // so pre-populate highest and lowest based on the first 2 items.
  // we could also start these as null and check below if they're set
  // but this is arguably cleaner
  var highest = Math.max(arrayOfInts[0], arrayOfInts[1]);
  var lowest  = Math.min(arrayOfInts[0], arrayOfInts[1]);

  var highestProductOf2 = arrayOfInts[0] * arrayOfInts[1];
  var lowestProductOf2  = arrayOfInts[0] * arrayOfInts[1];

  // except this one--we pre-populate it for the first *3* items.
  //this means in our first pass it'll check against itself, which is fine.
  var highestProductOf3 = arrayOfInts[0] * arrayOfInts[1] * arrayOfInts[2];

  // walk through items, starting at index 2
  for (var i = 2; i < arrayOfInts.length; i++) {
    var current = arrayOfInts[i];

    // do we have a new highest product of 3?
    // it's either the current highest,
    // or the current times the highest product of two
    // or the current times the lowest product of two
    highestProductOf3 = Math.max(Math.max(
      highestProductOf3,
      current * highestProductOf2),
      current * lowestProductOf2);

    // do we have a new highest product of two?
    highestProductOf2 = Math.max(Math.max(
      highestProductOf2,
      current * highest),
      current * lowest);

    // do we have a new lowest product of two?
    lowestProductOf2 = Math.min(Math.min(
      lowestProductOf2,
      current * highest),
      current * lowest);

    // do we have a new highest?
    highest = Math.max(highest, current);

    // do we have a new lowest?
    lowest = Math.min(lowest, current);
  }

  return highestProductOf3;
}