javascript - codewars一道推理题，不理解，题名字是 Unflatten a list

Question

• You have to do several runs. The depth is the number of runs, you have to do.

• In every run you have to switch the direction. First run from left, next run from right. Next left...
  Every run has these rules:

• You start at the first number (from the direction).

• Take for every number x the remainder of the pision by the number of still available elements (from
  this position!) to have the number for the next decision.

• If the remainder-value is smaller than 3, take this number x (NOT the remainder-Value) direct
  for the new array and continue with the next number.

• If the remainder-value (e.g. 3) is greater than 2, take the next remainder-value-number (e.g. 3)
  elements/numbers (inclusive the number x, NOT the remainder-value) as a sub-array in the new array.

Continue with the next number/element AFTER this taken elements/numbers.

• Every sub-array in the array is independent and is only one element for the progress on the array.
  For every sub-array you have to follow the same rules for unflatten it.

The direction is always the same as the actual run.

Array: [4, 5, 1, 7, 1] Depth: 2 -> [[ 4, [ 5, 1, 7 ] ], 1]

Steps:
First run: (start from left side!)

1. The first number is 4. The number is smaller than the number of remaining elements, so it is the remainder-value (4 / 5 -> remainder 4).
   So 4 numbers (4, 5, 1, 7) are added as sub-array in the new array.

2. The next number is 1. It is smaller than 3, so the 1 is added direct to the new array.
   Now we have --> [[4, 5, 1, 7], 1]

Second run: (start from right side!)

1. The last number (first from other side) is 1. So the 1 is added direct to the new array.

2. The next element is the sub-array. So we use the rules for this.
   2a.The last number is 7. There are 4 elements in the array. So for the next decision you have to

take the remainder from 7 / 4 -> 3. So 3 numbers (5, 1, 7) are added as sub-array in the
new array.
2b.Now there is the 4 and only one element last in this array. 4 / 1 -> remainder 0. It is smaller
than 3. So the 4 is added direct to the new array.
Now we have --> [[ 4, [ 5, 1, 7 ] ], 1]