# Make change for a dollar amount

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Scott am 5 Dez. 2014
I need to write a function that generates all of the ways that change for a an amount can be made. However, we are not using nickle, dimes, and quarters to make this change. We are using a set of inputted values.
For example, if my inputs are: X = [2 3 4] and Y = 32
My output should be:
16 0 0
14 0 1
10 0 3
etc...,
showing all of the ways that I can make a sum of 32 with the coin values 2,3, and 4.
I've tried to use some sort of permutation to do this, and I've also tried using mod(x,y) in different fashions, and cannot seem to get the correct result.
Any insights? Thanks a lot folks!
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### Akzeptierte Antwort

Mohammad Abouali am 6 Dez. 2014
Bearbeitet: Mohammad Abouali am 6 Dez. 2014
This question usually comes out a lot in job-interviews. The common method of implementation is using a recursive function. Save the following function in payBill.m
function payBill(billList,billCount,Amount)
billList=sort(billList,'descend');
maxN=floor(Amount/billList(1));
if (numel(billList)==1)
reminder=mod(Amount,billList);
if (reminder==0)
count=[billCount maxN];
disp(count)
end
elseif (numel(billList)>1)
for i=maxN:-1:0
reminder=Amount-i*billList(1);
payBill(billList(2:end),[billCount i],reminder);
end
end
end
Now let's say you want to see all the combination of paying 7 cents using 1cent, 2cent, and 5cent coins. You can type the following command
payBill([1 5 2],[],7)
It will generate all the combination possible as follow:
1 1 0 % 1x 5 cents + 1x 2 cents
1 0 2 % 1x 5 cents + 2x 1 cents
0 3 1 % 3x 2 cents + 1x 1 cents
0 2 3 % 2x 2 cents + 3x 1 cents
0 1 5 % 1x 2 cents + 5x 1 cents
0 0 7 % 7x 1 cents
Note 1: when calling the function the order of billList doesn't matter. We could have put in [1 2 5] or [2 5 1] or [1 5 2], doesn't matter.
Note 2: when the counts are printed, regardless of the order you put the billList it would be always from the highest value coin to the lowest. Note in our example we gave [1 5 2] but the first number in the output is the count of 5 cents coin.
Note 3: if there is no possible combination to pay the exact amount nothing is printed. try payBill([3 5],[],7). There is no combination of 3cents coin and 5 cents coin to pay 7 cents. so nothing would be printed.
to get your example type, i.e. paying 32 cents using 2cents, 3cents, and 4cents coin type:
payBill([2,3,4],[],32)
the following output is generated. Note that the first column is for the highest value coin, i.e. 4cents and then the next column is for 3 cents and the last column is for 2cents.
8 0 0 % 8x 4 cents
7 0 2 % 7x 4 cents + 2x 2 cents
6 2 1 % and so on
6 0 4
5 4 0
5 2 3
5 0 6
4 4 2
4 2 5
4 0 8
3 6 1
3 4 4
3 2 7
3 0 10
2 8 0
2 6 3
2 4 6
2 2 9
2 0 12
1 8 2
1 6 5
1 4 8
1 2 11
1 0 14
0 10 1
0 8 4
0 6 7
0 4 10
0 2 13
0 0 16
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Mohammad Abouali am 6 Dez. 2014
Bearbeitet: Mohammad Abouali am 6 Dez. 2014
Nope. Doesn't need to change anything in the code. The code already takes care of that
If you look at note1 in my first reply as I said, doesn't matter how the denominations are inputed. It can be ascending, descending or even random order.
The output would be always in the descending order (highest denomination to the lowest)
If you want the output to be also in ascending order just flip the results using fliplr() , as follow:
count=fliplr( payBill([2:4],32) );
Mohammad Abouali am 6 Dez. 2014
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