This will help us find the proper decrease function We decrease the amount of quarters by one this wayДef decreaseQuarter(base: Array): Array =Дef decreaseDime(base: Array): Array =Дef decreaseNickel(base: Array): Array = few types of coins, it's not worth it. be made to receive a list of adjustments, but for so adding or subtracting coins of each type this could RemainingAmount :: numberOfCoins :: list.tail Val remainingAmount = currentAmount % coinValue Val numberOfCoins = currentAmount / coinValue

Val coinNames = List(('quarter', 'quarters'),ДЇunction.tupled((quarters: Int, dimes: Int, nickels:Int, pennies: Int) => ( A simple solution would be this: // Generate a pretty string Below there is an overly complicated solution, which I'll preserve because it is a solution, after all. If (i < 0 || j < 0) // Impossible, for illustration purposeД®lse if (i = 0 || j = 0) // Very Important (quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent) */ДЇ(n, k): number of ways of making change for n cents, using only the first There are only penny, nickel, dime, and quarter. 200 = 2 dollars, 1000 = 10 dollars),ДЇind the number of combinations of coins that make up the dollar value. * Q: Given some dollar value in cents (e.g. The sub problem is a typical Dynamic Programming problem. Return sum(count_combs(left-x*cur, i+1, comb, (x,cur)) for x in range(0, int(left/cur)+1)) You can use 0 to k copies of your current denomination such that k * cur denomination 0:Д¬omb.append( (left/denominations, demoninations) )

If there is only 1 denomination left, there is only one way to fill the total. You have a current total to fill, and a largest denomination (with more than 1 left).You have some list of denominations, if the smallest one can evenly divide any remaining currency amount, this should work fine.Д«asically, you move from largest to smallest denominations.

All us coins