Make change for a dollar amount

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!

Answer 1

[Mohammad Abouali]

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

[Scott]

That appears to do exactly what I need! thank you so much.

But if i only want the function to take two inputs, namely:

1) the denominations of the change that can be made i.e. [1 3 5] 2) the total amount that I would like to make change for i.e. 30

[Mohammad Abouali]

So I changed the code in such a way that it does not require that third parameter at all any more. so only two parameters (those that you want)

Here is the new code:

function count=payBill(billList,Amount)
  billList=sort(billList,'descend');
  maxN=floor(Amount/billList(1));
  if (numel(billList)==1)
    reminder=mod(Amount,billList);
    if (reminder==0)
      count=[billList;maxN];
    else
      count=[];
    end
    return;
  elseif (numel(billList)>1)
    count=reshape(billList,1,[]);
    for i=maxN:-1:0
      reminder=Amount-i*billList(1);
      tmpCount=payBill(billList(2:end),reminder);
      count=[count; ...
             ones(size(tmpCount,1)-1,1)*i tmpCount(2:end,:)];
    end
  end
end

Note that the method to call the function has changed now. It actually returns all the combination and the first row is the ordered bill list (denominations). Here is an example:

count=payBill([1,2,5],5)

count =

       5     2     1 % Note the first row is the denominations 
                     % to guid each column coincide with which coin
       1     0     0
       0     2     1
       0     1     3
       0     0     5

If this is what you want, please accept the answer.

[Scott]

well done sir! exactly what i was looking for.

now if I wanted each of those triplets, i.e. 0 1 3 0 0 5

to be outputted each as it's own cell array, how could i accomplish that? so the desired output is: 1 3, 0 5, etc...

[Mohammad Abouali]

Oh this last one is easy. Use mat2cell() function as follow:

count=payBill([2:4],32)
countCell=mat2cell(count,ones(size(count,1),1),size(count,2))

So countCell is gonna be cell array. each element is one triplet of count or generally one row of count. You can retrieve them later as follow

>> countCell{1} % note the first one is the denominations though.

ans =

       4     3     2

>> countCell{2}

ans =

       8     0     0

[Scott]

and actually it is assumed that our change denominations will be given in an ascending order. so that will change the code right?

[Mohammad Abouali]

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]

If your question is answered, please accept the answer.