algorithm – How to find all combinations of coins when given some dollar value – Stack Overflow

Duh, I feel stupid right now. Below there is an overly complicated solution, which I’ll preserve because it is a solution, after all. A simple solution would be this:

// Generate a pretty string
val coinNames = List(("quarter", "quarters"), 
                     ("dime", "dimes"), 
                     ("nickel", "nickels"), 
                     ("penny", "pennies"))
def coinsString = 
  Function.tupled((quarters: Int, dimes: Int, nickels:Int, pennies: Int) => (
    List(quarters, dimes, nickels, pennies) 
    zip coinNames // join with names
    map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
    map (t => t._1 + " " + t._2) // qty name
    mkString " "
  ))

def allCombinations(amount: Int) = 
 (for{quarters <- 0 to (amount / 25)
      dimes <- 0 to ((amount - 25*quarters) / 10)
      nickels <- 0 to ((amount - 25*quarters - 10*dimes) / 5)
  } yield (quarters, dimes, nickels, amount - 25*quarters - 10*dimes - 5*nickels)
 ) map coinsString mkString "\n"

Here is the other solution. This solution is based on the observation that each coin is a multiple of the others, so they can be represented in terms of them.

// Just to make things a bit more readable, as these routines will access
// arrays a lot
val coinValues = List(25, 10, 5, 1)
val coinNames = List(("quarter", "quarters"), 
                     ("dime", "dimes"), 
                     ("nickel", "nickels"), 
                     ("penny", "pennies"))
val List(quarter, dime, nickel, penny) = coinValues.indices.toList


// Find the combination that uses the least amount of coins
def leastCoins(amount: Int): Array[Int] =
  ((List(amount) /: coinValues) {(list, coinValue) =>
    val currentAmount = list.head
    val numberOfCoins = currentAmount / coinValue
    val remainingAmount = currentAmount % coinValue
    remainingAmount :: numberOfCoins :: list.tail
  }).tail.reverse.toArray

// Helper function. Adjust a certain amount of coins by
// adding or subtracting coins of each type; this could
// be made to receive a list of adjustments, but for so
// few types of coins, it's not worth it.
def adjust(base: Array[Int], 
           quarters: Int, 
           dimes: Int, 
           nickels: Int, 
           pennies: Int): Array[Int] =
  Array(base(quarter) + quarters, 
        base(dime) + dimes, 
        base(nickel) + nickels, 
        base(penny) + pennies)

// We decrease the amount of quarters by one this way
def decreaseQuarter(base: Array[Int]): Array[Int] =
  adjust(base, -1, +2, +1, 0)

// Dimes are decreased this way
def decreaseDime(base: Array[Int]): Array[Int] =
  adjust(base, 0, -1, +2, 0)

// And here is how we decrease Nickels
def decreaseNickel(base: Array[Int]): Array[Int] =
  adjust(base, 0, 0, -1, +5)

// This will help us find the proper decrease function
val decrease = Map(quarter -> decreaseQuarter _,
                   dime -> decreaseDime _,
                   nickel -> decreaseNickel _)

// Given a base amount of coins of each type, and the type of coin,
// we'll produce a list of coin amounts for each quantity of that particular
// coin type, up to the "base" amount
def coinSpan(base: Array[Int], whichCoin: Int) = 
  (List(base) /: (0 until base(whichCoin)).toList) { (list, _) =>
    decrease(whichCoin)(list.head) :: list
  }

// Generate a pretty string
def coinsString(base: Array[Int]) = (
  base 
  zip coinNames // join with names
  map (t => (if (t._1 != 1) (t._1, t._2._2) else (t._1, t._2._1))) // correct for number
  map (t => t._1 + " " + t._2)
  mkString " "
)

// So, get a base amount, compute a list for all quarters variations of that base,
// then, for each combination, compute all variations of dimes, and then repeat
// for all variations of nickels.
def allCombinations(amount: Int) = {
  val base = leastCoins(amount)
  val allQuarters = coinSpan(base, quarter)
  val allDimes = allQuarters flatMap (base => coinSpan(base, dime))
  val allNickels = allDimes flatMap (base => coinSpan(base, nickel))
  allNickels map coinsString mkString "\n"
}

So, for 37 coins, for example:

scala> println(allCombinations(37))
0 quarter 0 dimes 0 nickels 37 pennies
0 quarter 0 dimes 1 nickel 32 pennies
0 quarter 0 dimes 2 nickels 27 pennies
0 quarter 0 dimes 3 nickels 22 pennies
0 quarter 0 dimes 4 nickels 17 pennies
0 quarter 0 dimes 5 nickels 12 pennies
0 quarter 0 dimes 6 nickels 7 pennies
0 quarter 0 dimes 7 nickels 2 pennies
0 quarter 1 dime 0 nickels 27 pennies
0 quarter 1 dime 1 nickel 22 pennies
0 quarter 1 dime 2 nickels 17 pennies
0 quarter 1 dime 3 nickels 12 pennies
0 quarter 1 dime 4 nickels 7 pennies
0 quarter 1 dime 5 nickels 2 pennies
0 quarter 2 dimes 0 nickels 17 pennies
0 quarter 2 dimes 1 nickel 12 pennies
0 quarter 2 dimes 2 nickels 7 pennies
0 quarter 2 dimes 3 nickels 2 pennies
0 quarter 3 dimes 0 nickels 7 pennies
0 quarter 3 dimes 1 nickel 2 pennies
1 quarter 0 dimes 0 nickels 12 pennies
1 quarter 0 dimes 1 nickel 7 pennies
1 quarter 0 dimes 2 nickels 2 pennies
1 quarter 1 dime 0 nickels 2 pennies

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