coin 2

coin2

cases when the amount cannot be made up:

amount < min-denomination

smallest-remainder-of-amount / min-denom is not an integer

actually the above isn't right - if the coins were say 71, 4 and 3 and the amount was 75

Actually in that case smallest-remainder-of-amount would be 0 

0/x is still 0

is 0 an integer?

https://www.google.co.uk/search?q=is+zero+an+integer&rlz=1C1CHBF_en-GBGB741GB742&oq=is+zero+an+in&aqs=chrome.2.69i57j6j0l4.3728j1j7&sourceid=chrome&ie=UTF-8
https://www.quora.com/Is-0-an-integer

it would appear so

so in order to work out the cases where we return -1 we need to:

a) calculate smallest-remainder-of-amount
b) have an "integer checker"

we can then use this to calculate our answer as well because "smallest-remainder of amount" is 0 we have counted through amount

would also need a "counter" keeping track of how many of each coin we have used to get down from amount to 0

our denominations range from denom-1 to denom-max each of which have an actual numerical value

we want to take amount, subtract denom-max until the remainder is smaller than 1 * denom-max, then apply the same process with denom-(max - 1)

until we have an amount that is 0

if we arrive at an amount smaller than the current deonomination AND smaller than the minimum denomination we stop and return -1

define coin-iter a denom-x
    (- amount (* a denom-x)
               denom-x)))
               
if ( < (coin-iter a denom-x) denom-x)
and (> (coin-iter a denom-x) 0))
then (coin-iter a denom-(- x 1)
else  (coin-iter (- a 1) denom-x)