Solving an 0 1 Knapsack Problem with LAMBDA

Patrick2788  So I went with the recursive call option:

MaxPac = LAMBDA(items, benefits, weights, limit, [hide_text], LET(
    i, IF(ROWS(items) = 1, TRANSPOSE(items), items),
    b, IF(ROWS(benefits) = 1, TRANSPOSE(benefits), benefits),
    w, IF(ROWS(weights) = 1, TRANSPOSE(weights), weights),
    list, REDUCE(0, SEQUENCE(ROWS(i)), LAMBDA(q, p,
             IF(INDEX(w, p) <= limit,
               VSTACK(q, IF(ROWS(i) > p,
                  LET(sublist, MaxPac(DROP(i, p),DROP(b, p),DROP(w, p),limit - INDEX(w, p)),
                      IF( IFERROR(INDEX(sublist, 1), 0),
                         HSTACK(INDEX(b, p), INDEX(i, p)),
                         HSTACK(INDEX(b, p) + INDEX(sublist, 1, 1),
                                INDEX(i, p),
                                DROP(sublist, , 1))
                         )),
                  HSTACK(INDEX(b, p), INDEX(i, p))
                  )),
            q))),
    out, INDEX(SORT(list, 1, -1), 1, ),
    hideText, IF(ISOMITTED(hide_text), FALSE, hide_text),
    IF( hideText, out,
        CONCAT("Benefit: ", INDEX(out, 1), "  Items: ",
            IFERROR(TEXTJOIN(",", 1, DROP(out, , 1)), 0)
        )
    )
))

It looks complicated but in concept it is easy.  I check loop using REDUCE and check if that element is less than the weight limit, if so then I 'add' it to the list and re-call itself with the remainder of the elements.  Each time I add a row and each row sums the Benefit in the first element.  The last step is to sort by the Benefit and only Take/Return the highest value and pass that up the chain.  There are a number of extra IFs and such because of making sure I don't strip off all the elements and such and in the beginning I check if the array is in a row or a column and swap it to make then columns if needed.

EDIT: I found some bug where the subcall must return an error or something in certain cases and I added an IFERROR() to the check.  I also added some text/formatting that will be added only on the top call if that parameter is missing or set to false.

So I found this to be significantly faster in cases when all 5 objects are selected (i.e. the weights are all set below the total limit).  I  can only imagine that additional item list would also result in time improvements also.

On final note, this solution intentionally did not attempt to return all potential best solutions and just returns a best solution.  In another words if there are more than 1 solution that returns the same maximum benefit, this solution will just return one of them.