True, in most cases the recursive form is slower since it is repeating many calculations (somewhat analogous to a recursive fibonnacci number calculation), whereas the iterative version following the 'bottom-up' calculation is referring back to a list being constructed for each amount up to the desired amount, possible since this problem exhibits optimal substructure - i.e. The optimal coins to make an amount 'n' is an optimal coin 'c0' + the optimal coins to make an amount 'n-c0' ...