Minimum Coin Exchange

David Wagner

In this post I solve the Minimum Coin Exchange problem programmatically using Haskell. I will compare the performance of the naive implementation to that using dynamic programming.

The problem

The minimum coin exchange problem is generally asked: if we want to make change for \(N\) cents, and we have infinite supply of each of \(S=\{S_{0},S_{2},\ldots ,S_{m-1}\}\) valued coins, what is the least amount of coins we need to make the change?

The solution can be found in a recursive manner where at each step of the recursion we have two options:

  1. we use the \(S_m\) valued coin in the change and we try to find the change for \(N - S_m\) cents with the same set of coins.

  2. we decide not to use the \(S_m\) valued coin in the change: we keep on looking for a change of \(N\) cents with \(m-1\) coins

At each step we need to choose the option that uses less coins. It is expected that this process will end after finite number of steps because either the number of coins or the money to change decreases at each step.

On the we find a succinct recursive formula to describe this process

$$ C(N,m)=\min {(C(N - S_{m}, m) + 1, C(N, m-1))} $$

the arguments of the \(\min\) function correspond to the two options described above. The \(+1\) in the first argument shows that choosing that option will increase the number of coins in the change. The recursive formula is completed with the base cases:

$$ \begin{align} C(N,m) &= 0, & N=0 \\ C(N,m) &= \infty, & N<0 \\ C(N,m) &= \infty, & N\geq 1, m\leq 0 \\ \end{align} $$

If the result of \(C(N,m)\) is either \(0\) or \(\infty\) then it is impossible make change for \(N\) with the given coins. The case (1) applies when we successfully changed \(N\) with the given coins. Case (2) means that our smallest valued coin is larger than the amount for which the change is requested. Finally (3) represents the situation where we cannot pick any more coin to complete the change.

For simplicity, we will use a fixed set of coins: \([25, 20, 10, 5, 1]\). For example, the minimum number of coins needed for 40 cents using the formulation above is given by \(C(40, 4)\). Also note that we only compute the number of coins, we don’t give the exact denominators to be used in the change.


As we use a fixed set of coins we hard-code the available coin denominators:

-- Available coin denominators.
coins :: [Int]
coins = [25, 20, 10, 5, 1]

We will write a function with the following signature:

change :: Int     -- amount
       -> Int     -- index of the last available coin
       -> Change  -- the number of coins

change takes to arguments: the amount to change, the index of the last available coin and it returns the number of coins in the change.

Taking a list index as the second argument is, of course, very error prone: our program will crash if we try to address an element that is not present in the coins list. We ignore this deficency in order to stay as close as possible to the theoretical formulation of the problem.

Since there are cases where the change is impossible, we choose the result type Change as

type Change = Maybe Int

This is more expressive than using infinity as a sentinel value when the change is not possible.

With this preparation we are ready to implement the first version of change.

Naive implementation

-- Naive implementation using the recursive formula from:
change :: Int -> Int -> Change
change n m
  | n == 0            = Just 0
  | n < 0             = Nothing
  | n >= 1 && m <= 0  = Nothing
  | otherwise         = minOf left right

    left = (+1) `fmap` change (n - sm) m
    right = change n (m - 1)
    sm = coins !! m

This implementation is almost maps one-to-one to the recursive formula above. The first three guard expression handle the three base cases. In the last clause the function calls itself to solve the two sub-problems which I called left and right.

Choosing Maybe to represent the result Change forces us to deviate from the clean mathematical formulation at two places:

  1. we use fmap to add 1 to the result of the left subproblem
  2. we use our own minOf function instead of the built-in min function

The first point is easy: we cannot add an Int to a Maybe Int because their types don’t match. Since Maybe Int is a functor so we can use fmap to lift the addition into the context of Maybe.

As for the second point, let’s see the implementation of minOf:

import Control.Applicative ((<|>))

minOf :: Change -> Change -> Change
minOf (Just i) (Just j) = Just (min i j)
minOf c1 c2             = c1 <|> c2

The function takes two Change values: if both Maybe contain Int values we choose the smaller one. Otherwise we try to keep the one that has a value in it using <|>. This behaves similarly to logical OR. We can easily test this in ghci:

Prelude> import Control.Applicative
Prelude Control.Applicative> Just 1 <|> Nothing  -- keeps the first
Just 1
Prelude Control.Applicative> Nothing <|> Just 2  -- keeps the second
Just 2
Prelude Control.Applicative> Nothing <|> Nothing -- returns Nothing
Prelude Control.Applicative> Just 1 <|> Just 2   -- prefers the first
Just 1

This implementation of minOf gives us the right behavior when we select the smaller between the results of the left and right subproblems.

The built-in min can actually operate on Maybe Int values. We just cannot use it here because it returns Nothing if any of its argument is Nothing. This would terminate our recursive function without exploring the whole solution space.

We could almost directly implement the terse mathematical solution as a recursive function. Overall our function is short and readable, but before we open the champagne and celebrate let’s see how our solution performs.

Performance of the naive solution

We can use the microbenchmarking library criterion to measure the running time of the naive change implementation. Let’s see how the running time depends on the amount to change. The following table shows the approximate time of computing the change for 40, 100, 150 and 200 cents.

Amount [cents] Running time [ms]
40 0.026
100 0.344
150 1.420
200 4.025

The running time of the naive solution scales polynomially with the number of cents. Let’s try to improve this!

Implementation using dynamic programming

The recursive method of the minimum coin exchange problem combines the solutions of subproblems with smaller amounts to change. We could optimize our naive solution using dynamic programming. The idea is that every time we solve a subproblem we memorize its solution. The next time the same subproblem occurs, instead of recomputing its solution we look up the previously solved solution.

We write a new function changeD which represents a stateful computation. The state is a map from problem parameters to its solution. In our case, a map from pair of integers (the denominator index and the amount) to a Change value. We call this computation Dyn:

import qualified Data.Map.Strict as M
import Control.Monad.Trans.State

type Dyn = State (M.Map (Int, Int) Change)

Using this, we can write changeD which, returns a Dyn computation resulting in a Change:

changeD :: Int -> Int -> Dyn Change
changeD n m
  | n == 0            = return $ Just 0
  | n < 0             = return Nothing
  | n >= 1 && m <= 0  = return Nothing
  | otherwise         = do

    left  <- memorize n (m - 1)
    right <- memorize (n - sm) m
    return $ minOf left (fmap (+1) right)

      sm = coins !! m

The code looks very much like the naive solution but, since we’re operating in the Dyn context, we are using the do notation. The memorize computation implements the storing and recalling the subproblems’ solution:

memorize :: Int -> Int -> Dyn Change
memorize n m = do
    val <- do
        elem <- gets $ M.lookup (n, m)  -- try to recall the solution
        case elem of
            Just x  -> return x         -- return previously stored solution
            Nothing -> changeD n m      -- compute the solution

    modify $ M.insert (n, m) val        -- store the solution
    return val

This function is a literal implementation of the dynamic programming method. We try to look up the solution in the state: if the subproblem has already been solved we return the solution, otherwise we solve the subproblem and store its solution in the state.

The final step is to provide change in a form identical to the naive solution:

change :: Int -> Int -> Change
change m n = evalState (changeD m n) M.empty

We execute the Dyn computation using evalState by providing an initial empty state. This function provides exactly the same interface as the naive version above. The two implementations can be used interchangeably. Let’s which of the two implementations is worth using.

Naive vs dynamic

The table below compares the running times of the two implementations as a function of the amount to change.

Amount [cents] Running time (naive) [ms] Running time (dynamic) [ms]
40 0.026 0.117
100 0.344 0.354
150 1.420 0.556
200 4.025 0.814

The dynamic programming version scales linearly with the amount to change. The difference in running time becomes significant for amounts larger than 100 cents. As always, this speedup didn’t come for free: we traded running speed for storage space.


The minimum coin exchange problem is a classic example demonstrating dynamic programming. We implemented a solution by naively transcribing the proposed recursive formula almost literally to Haskell. We then used dynamic programming to improve time complexity of the naive solution. The dynamic programming method was encapsulated in a Dyn computation where the solutions to already solved subproblems are stored in a map.

The code for both implementations can be found here.