9/13/2023 0 Comments American coins names and valuesThe dollar ( currency code USD and 840) is the unit of currency of the United States. We traverse from row money 1 to row money 25 (4 rows).The United States, the British Virgin Islands, East Timor, Ecuador, El Salvador, the Marshall Islands, Micronesia, Palau, Panama, Turks and Caicos Islands, and the insular areas of the United StatesĪWG, BSD, BHD, BBD, BSD, BMD, KYD, CUC, DJF, XCD, ERN, JOD, LBP, MVR, ANG, OMR, QAR, SAR, AED This is a dynamic programming solution with order(n) time. we'll produce a list of coin amounts for each quantity of that particularĭef coinSpan(base: Array, whichCoin: Int) = Given a base amount of coins of each type, and the type of coin, Val decrease = Map(quarter -> decreaseQuarter _, This will help us find the proper decrease function We decrease the amount of quarters by one this wayĭef decreaseQuarter(base: Array): Array =ĭef decreaseDime(base: Array): Array =ĭef 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"),įunction.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 purposeĮlse if (i = 0 || j = 0) // Very Important (quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent) */į(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),įind 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:Ĭomb.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.īasically, you move from largest to smallest denominations.
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