Refractor Diff. Two Arrays Code

Curriculum Help
1

I recently finished the diff. two arrays challenge in the intermediate scripting algorithms section of the JS certification and was wondering if my solution can be slimmed down a bit.

function check(cArr1, cArr2) {
  // First, map a new array from arr1 containing values not found in arr2...
  let filterArr1 = cArr1.map(
    function(a) {
      if (cArr2.includes(a) == false) {
        return a;
      }
    }
    // Then filter out any [""] values. These occur when the arrays are of different lengths.
  ).filter(a => a);
  // Do the same thing in reverse...
  let filterArr2 = cArr2.map(
    function(a) {
      if (cArr1.includes(a) == false) {
        return a;
      }
    }
  ).filter(a => a);
  return filterArr1.concat(filterArr2);
};

function diffArray(arr1, arr2) {
  return check(arr1, arr2);
}

console.log(diffArray([1, "calf", 3, "piglet"], [7, "filly"]));
3

It iterates though the second array’s values to find elements not present in the first array and maps them to a new array.

If I comment out that code block, the function will not be able to pass the tests when both arrays contain unique elements.

For example, if I remove that code block and call diffArray([1, 2, 3], [4, 5]), the returned value will just be [1, 2, 3] when the resulting array should be [1, 2, 3, 4, 5]. So I have to keep it there for now.

4

You don’t need the map method. You can simply use the filter method.

let filterArr1 = cArr1.filter(a => cArr2.includes(a) == false);

OR more concisely:

let filterArr1 = cArr1.filter(a => !cArr2.includes(a));
5

The diffArray function is unnecessary, you could just do this

console.log(check([1, "calf", 3, "piglet"], [7, "filly"]));