How to find all combinations of coins when given s

Question

I found a piece of code that I was writing for interview prep few months ago.

According to the comment I had, it was trying to solve this problem:

Given some dollar value in cents (e.g. 200 = 2 dollars, 1000 = 10 dollars), find all the combinations of coins that make up the dollar value. There are only pennies, nickels, dimes, and quarters allowed. (quarter = 25 cents, dime = 10 cents, nickel = 5 cents, penny = 1 cent)

For example, if 100 was given, the answer should be:

4 quarter(s) 0 dime(s) 0 nickel(s) 0 pennies  
3 quarter(s) 1 dime(s) 0 nickel(s) 15 pennies  
etc.

This can be solved in both iterative and recursive ways, I believe. My recursive solution is quite buggy, and I was wondering how other people would solve this problem. The difficult part of this problem was making it as efficient as possible.

Answer 1

I know this is a very old question. I was searching through the proper answer and couldn't find anything that is simple and satisfactory. Took me some time but was able to jot down something.

function denomination(coins, original_amount){
    var original_amount = original_amount;
    var original_best = [ ];

    for(var i=0;i<coins.length; i++){
      var amount = original_amount;
      var best = [ ];
      var tempBest = [ ]
      while(coins[i]<=amount){
        amount = amount - coins[i];
        best.push(coins[i]);
      }
      if(amount>0 && coins.length>1){
        tempBest = denomination(coins.slice(0,i).concat(coins.slice(i+1,coins.length)), amount);
        //best = best.concat(denomination(coins.splice(i,1), amount));
      }
      if(tempBest.length!=0 || (best.length!=0 && amount==0)){
        best = best.concat(tempBest);
        if(original_best.length==0 ){
          original_best = best
        }else if(original_best.length > best.length ){
          original_best = best;
        }  
      }
    }
    return original_best;  
  }
  denomination( [1,10,3,9] , 19 );

This is a javascript solution and uses recursion.

Answer 2

I looked into this once a long time ago, and you can read my little write-up on it. Here’s the Mathematica source.

By using generating functions, you can get a closed-form constant-time solution to the problem. Graham, Knuth, and Patashnik’s Concrete Mathematics is the book for this, and contains a fairly extensive discussion of the problem. Essentially you define a polynomial where the nth coefficient is the number of ways of making change for n dollars.

Pages 4-5 of the writeup show how you can use Mathematica (or any other convenient computer algebra system) to compute the answer for 10^10^6 dollars in a couple seconds in three lines of code.

(And this was long enough ago that that’s a couple of seconds on a 75Mhz Pentium...)

Answer 3

Both: iterate through all denominations from high to low, take one of denomination, subtract from requried total, then recurse on remainder (constraining avilable denominations to be equal or lower to current iteration value.)

Answer 4

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

Answer 5

I found this neat piece of code in the book "Python For Data Analysis" by O'reily. It uses lazy implementation and int comparison and i presume it can be modified for other denominations using decimals. Let me know how it works for you!

def make_change(amount, coins=[1, 5, 10, 25], hand=None):
 hand = [] if hand is None else hand
 if amount == 0:
 yield hand
 for coin in coins:
 # ensures we don't give too much change, and combinations are unique
 if coin > amount or (len(hand) > 0 and hand[-1] < coin):
 continue
 for result in make_change(amount - coin, coins=coins,
 hand=hand + [coin]):
 yield result

Answer 6

This is the improvement of Zihan's answer. The great deal of unnecessary loops comes when the denomination is just 1 cent.

It's intuitive and non-recursive.

    public static int Ways2PayNCents(int n)
    {
        int numberOfWays=0;
        int cent, nickel, dime, quarter;
        for (quarter = 0; quarter <= n/25; quarter++)
        {
            for (dime = 0; dime <= n/10; dime++)
            {
                for (nickel = 0; nickel <= n/5; nickel++)
                {
                    cent = n - (quarter * 25 + dime * 10 + nickel * 5);
                    if (cent >= 0)
                    {
                        numberOfWays += 1;
                        Console.WriteLine("{0},{1},{2},{3}", quarter, dime, nickel, cent);
                    }                   
                }
            }
        }
        return numberOfWays;            
    }