# Coin change logic

#### Tags: coin-change, java

Im stuck at this problem about the change of a vending machine (using 10ct, 20 ct, 50ct, 100ct and 200ct-coins.)

So lets say coffee costs 40cts. The user throws in 2€ (labeled 200cts).

Now im supposed to figure out how the change of 160cts is given back to the user. There are 2 conditions: A) Taking the shortest combination, but B) only if the register has enough coins to hand out said combination .

So in my example, the shortest combination is 100cts + 50cts + 10cts. But if, lets say, there are no 10ct coins left in the register, the prefered combination should be 100ct + 20ct + 20ct + 20ct.

```public void coinChange (int change) {

int TwoEuroCount = 0, OneEuroCount= 0, FiftyCentsCount = 0, TwentyCentsCount = 0, TenCentsCount = 0;

while (change > 0) {

TwoEuroCount = change / 200;

if(register.availableTwoEuros(TwoEuroCount) == true) {
register.withdrawTwoEuros(TwoEuroCount);
change = change - 200 * TwoEuroCount;

//the method .availableTwoEuros returns true if AmountOfTwoEuros - TwoEuroCount >= 0
}

OneEuroCount = change / 100;

if(register.availableOneEuro(OneEuroCount) == true) {
register.withdrawOneEuro(OneEuroCount);
change = change - 100 * OneEuroCount;
}

FiftyCentsCount = change / 50;

if(register.availableFiftyCents(FiftyCentsCount) == true) {
register.withdrawFiftyCents(FiftyCentsCount);
change = change - 50 * FiftyCentsCount;
}

TwentyCentsCount = change / 20;

if (register.availableTwentyCents(TwentyCentsCount) == true) {
register.withdrawTwentyCents(TwentyCentsCount);
change = change - 20 * TwentyCentsCount;
}

TenCentsCount = change / 10;

if(register.availableTenCents(TenCentsCount) == true) {
register.withdrawTenCents(TenCentsCount);
change = change - 10 * TenCentsCount;
}
}
}
```

This works perfectly for finding the shortest combination if there are enough coins. But if i start with AmountTenCents = 0, the method will just take 1 Euro and 50cts and leave it at that.

Suppose you have:

• an array of all possible coin `VALUES`: [10, 20, 50, 100, 200]
• an array of the current `SUPPLY` of coins for each `VALUE`
• an array of `WEIGHS` that correspond to `VALUES` (higher weigh, smaller value): [4, 3, 2, 1, 0]

then you could find a combination of coins that sums up to change and has the minimum total weight.

Let a combination `c` be the current combination of coins. For example, `c = [0, 1, 1, 2, 0]` would mean that you are considering a combination where you have no 10 cent coins, one 20 cent coin, one 50 cent coin, two 1€ coins and no 2€ coins.

You begin with combination `c = [0, 0, 0, 0, 0]`.

Using weights will implicitly assure you that the resulting combination will have the minimum weight and is thus the result you are looking for. For example:

```// Both combinations represent the change of 160 cents
c = [1, 0, 1, 1, 0] => weight: 4*1 + 3*0 + 1*2 + 1*1 + 0*0 = 7
c = [0, 3, 0, 1, 0] => weight: 4*0 + 3*3 + 0*2 + 1*1 + 0*0 = 10
```

Something like this should work:

```import java.util.Arrays;
import java.util.stream.IntStream;

public class Change {
/** The number of unique coins. */
static final int N = 5;
static final int[] VALUES = { 10, 20, 50, 100, 200 };
static final int[] WEIGHTS = { 4, 3, 2, 1, 0 };
static final int[] SUPPLY = { 10, 35, 40, 100, 2 };

static int[][] result = {
{
// The minimum weight
Integer.MAX_VALUE
},
{
// The resulting combination of coins
0, 0, 0, 0, 0
}
};

public static void main(String[] args) {
int change = 160;
solve(new int[N], change);

if (result == Integer.MAX_VALUE) {
System.out.println(
"Can't return the change with the given SUPPLY of coins"
);
} else {
System.out.println(Arrays.toString(result));
}
}

static void solve(int[] c, int change) {
// check if out of supply
boolean isOutOfSupply = IntStream.range(0, N).anyMatch(i -> SUPPLY[i] < c[i]);
if (isOutOfSupply) return;

// compute weight
int weight = IntStream.range(0, N).map(i -> WEIGHTS[i] * c[i]).sum();

// compute sum
int sum = IntStream.range(0, N).map(i -> VALUES[i] * c[i]).sum();

if (sum == change && weight < result) {
result = weight;
result = c;
} else if (sum < change) {
IntStream.range(0, N).forEach(i -> solve(increment(c, i), change));
}
}

static int[] increment(int[] array, int index) {
int[] clone = array.clone();
clone[index]++;
return clone;
}
}
```

Source: stackoverflow