JS how to sort() one array based on another array’s sortation

When A is .sorted(), it becomes 6, 7, 8, so those number get new index values, so do the same for B. get current index of the values inside B, then rearrange it based on A’s new arrangement. So when A = 6,7,8, B would be u, z, h

```var arrA = [8, 6, 7] // B follows this arr (A)
var arrB = ['h', 'u', 'z'] // B follows A
arrA.sort()
// output: 6, 7, 8
// arrB.followA’s arrangement somehow
// output: u, z, h
arrA.reverse()
// output: 8, 7, 6
// output: h, z, u
console.log(arrA);
console.log(arrB)
```

Create a two-dimensional working array, whose elements are pairs of values from `arrA` and `arrB`:

```var work = [];
arrA.forEach(function( v, i ) {
work[ i ] = [ arrA[i], arrB[i] ];
});
```

Now you can arrange `work` in any order, and the values from `arrA` and `arrB` will stay in lockstep:

```work.sort(function(x, y) {
return Math.sign( x[0], y[0] );
});
```

(In the above example, we are ordering by the element in slot 0, which is from `arrA`. To order by the elements in `arrB`, change 0 to 1.)

Now you can alter `work`, e.g.:

```work.reverse();
```

And extract the corresponding elements that were originally from `arrA`:

```let newAarrA = work.map(function(x) {
return x[0]; // from arrA
});
console.log(newArrA);
```

(Change 0 to 1 to get the corresponding elements from `arrB` instead.)