Given string S representing a dollar amount, make change for that amount of money use the least number of coins to make the change and record the amount of each coin in a list. Here are the coins available to use and their value.

Coin : Value
Dollar Coins : $1.00
Quarters: $0.25
Dimes: $0.10
Nickels: $0.05
Pennies: $0.01


String S that contains the dollar symbol $ and the dollar amount.


List of coin numbers separated by a space character " ". The list must be in this order: Dollar coins, quarters, dimes, nickels, pennies.


  • $0.00 < S < $10.00
  • S is given to two decimal places.
  • make change for that amount of money use the least number of coins

Example Input




4 2 0 1 3

9 3 2 0 4

Win Condition

shortest bytes win.

  • 4
    \$\begingroup\$ Do we have to get the $ sign in the input? I suggest allowing the number itself, either as a number or a string (optional). \$\endgroup\$ Apr 16, 2018 at 18:48
  • 1
    \$\begingroup\$ Knapsack problem, possible duplicate of codegolf.stackexchange.com/questions/162158/knapsack-problem \$\endgroup\$
    – Angs
    Apr 16, 2018 at 18:57
  • 2
    \$\begingroup\$ @0x45 I'd upvote this if you remove the $-sign requirement in the input, and make the output format optional... \$\endgroup\$ Apr 16, 2018 at 19:03
  • 2
    \$\begingroup\$ Fair enough... You might want to read this. \$\endgroup\$ Apr 16, 2018 at 19:16
  • 2
    \$\begingroup\$ On further thought, this input format isn't interesting either. Because the format is always $d.dd, the symbols give no information and so just need to be stripped away or ignored. \$\endgroup\$
    – xnor
    Apr 16, 2018 at 23:39

9 Answers 9


Hexagony, 35 32 28 27 bytes


Try it online!

There are no no-ops . now, so it must be optimal.


  • \$\begingroup\$ -1 byte. This way of using "functions" in Hexagony is very interesting \$\endgroup\$
    – Jo King
    Apr 17, 2018 at 13:20
  • \$\begingroup\$ @JoKing Thanks! (originally I intend to use $ or > and throw it in a loop, but then figure out this way after trying to rotate the program to save some bytes) \$\endgroup\$
    Apr 17, 2018 at 13:26
  • \$\begingroup\$ One more. This one ends in a divide by zero error and prints an excess space though (also, what program are using to create the images? It doesn't look like the one I usually use) \$\endgroup\$
    – Jo King
    Apr 17, 2018 at 13:31
  • \$\begingroup\$ @JoKing Hexagony Colorer (of course), but I have to remove the caps (and I was too lazy to reimplement it) because of a Mono bug. (also with some change-font feature...) \$\endgroup\$
    Apr 18, 2018 at 14:23

Python 2, 79 72 71 bytes

for c in(100,25,10,5,1):print d/c,;d-=d/c*c

Try it online!

Saved 7 bytes thanks, in part, to Chas Brown

Saved a byte thanks to Jo King

  • \$\begingroup\$ Save 6 via s=input();d=int(s[1]+s[-2:]). \$\endgroup\$
    – Chas Brown
    Apr 17, 2018 at 3:13
  • \$\begingroup\$ Thanks, I totally overlooked that the input was less than $10, though my previous was flexible. :$ \$\endgroup\$ Apr 17, 2018 at 15:03
  • \$\begingroup\$ What's the ~~ for? \$\endgroup\$
    – Jo King
    Apr 19, 2018 at 21:27
  • \$\begingroup\$ @JoKing The ~ does a bitwise two's complement, and two of them reverses it, but essentially performs a Math.floor in many languages. And now I realize that the / operator does just that in Python 2 in this case. Thanks for pointing that out! Saved a byte. \$\endgroup\$ Apr 20, 2018 at 13:53
  • \$\begingroup\$ suggest d%=c instead of d-=d/c*c \$\endgroup\$
    – c--
    Jul 21, 2022 at 0:39

05AB1E, 24 17 bytes

¦'.¡25‰10‰5‰r5F?' ?}


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My first attempt at it. It's very verbose in my opinion and I'll look at further revising this answer. Maybe somebody better than me will figure it out.


¦             # Remove '$' from input
'.¡           # Push '.' and split string on that. (dollar),(cents) is result
`25‰          # Push all items onto stack and mod top by 25
`T‰           # Push all items onto stack and mod top by 10
`5‰           # Push all items onto stack and mod top by 5
`r            # Push all items onto stack and reverse stack
ðý            # Loop through stack and print each element with space

-7 bytes thanks to Emigna

  • 1
    \$\begingroup\$ Some quick gains: Replace 10 with T and r5F?' ?} with ðý. \$\endgroup\$
    – Emigna
    Apr 16, 2018 at 19:55
  • \$\begingroup\$ @Emigna How did you go about translating the r5F?' ?} to ðý if you don't mind so that I can know for future challenges. \$\endgroup\$
    – Multi
    Apr 16, 2018 at 20:00
  • 1
    \$\begingroup\$ ðý means join by space, which gives the same result as the print each with a space after that you were doing (except with a trailing newline instead of a trailing space) \$\endgroup\$
    – Emigna
    Apr 16, 2018 at 20:19
  • \$\begingroup\$ @Emigna Oh! I see now. I thought i was some sort of weird encoding, but it is literally push space and b.join(a). I got it now. Thanks :) \$\endgroup\$
    – Multi
    Apr 16, 2018 at 21:01

Octave, 106 bytes


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16 bytes is used simply to get rid of $ and convert the string to a number. The rest is simply removing the as many coins as possible, one value at a time.


R, 81 bytes

x=as.double(substr(scan(,""),2,6));for(i in 1/c(1,4,10,20,100))x=x-i*print(x%/%i)

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26 bytes ensuring the data are a double rather than a string, very nearly a third of the program! That's lame, to say the least.


Stax, 18 17 bytes


Run and debug it


Ruby, 74 bytes

->s{n=s.delete("$.").to_i;["%s"]*5*" "%[n/100,n/25%4,(m=n/5%5)/2,m%2,n%5]}

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Converts input to an integer containing number of pennies, then applies formulas in the [] at the end.

5 bytes standard Ruby boilerplate ->s{}
34 bytes on solving the problem
35 bytes on formatting

Julia, 72 71 bytes


Attempt This Online!

  • -1 byte: x=x%u -> x*=u - thanks to @MarcMush

Java 8, 138 bytes

s->{long p=new Long(s.replaceAll("\\D","")),c,i=100;for(s="";i>0;s+=c+" ",i=i>25?25:i>10?10:i>5?5:i>1?1:0)for(c=0;p>=i;p-=i)c++;return s;}


Try it online.

s->{              // Method with String as both parameter and return-type
  long p=new Long(s.replaceAll("\\D","")),
                  //  Remove all non-digits, and convert it to a number
       c,         //  Count-integer
       i=100;     //  Amount integer, starting at 100
  for(s="";       //  Set the input to an empty String, because we no longer need it
      i>0         //  Loop as long as `i` is not 0
      ;           //    After every iteration:
       s+=c+" ",  //     Append the count and a space to `s`
       i=i>25?    //     If `i` is 100:
          25      //      Change it to 25
         :i>10?   //     Else-if `i` is 25:
          10      //      Change it to 10
         :i>5?    //     Else-if `i` is 10:
          5       //      Change it to 5
         :i>1?    //     Else-if `i` is 5:
          1       //      Change it to 1
         :        //     Else (`i` is 1)
          0)      //      Change it to 0
    for(c=0;      //   Reset the count `c` to 0
        p>=i;     //   Inner loop as long as `p` is larger than or equal to `i`
        p-=i)     //     After every iteration: subtract `i` from `p`
      c++;        //    Increase the count `c` by 1
  return s;}      //  Return the result

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