Count the coins

From Rosetta Code
Task
Count the coins
You are encouraged to solve this task according to the task description, using any language you may know.

There are four types of common coins in   US   currency:

  1.   quarters   (25 cents)
  2.   dimes   (10 cents)
  3.   nickels   (5 cents),   and
  4.   pennies   (1 cent)


There are six ways to make change for 15 cents:

  1.   A dime and a nickel
  2.   A dime and 5 pennies
  3.   3 nickels
  4.   2 nickels and 5 pennies
  5.   A nickel and 10 pennies
  6.   15 pennies


Task

How many ways are there to make change for a dollar using these common coins?     (1 dollar = 100 cents).


Optional

Less common are dollar coins (100 cents);   and very rare are half dollars (50 cents).   With the addition of these two coins, how many ways are there to make change for $1000?

(Note:   the answer is larger than   232).


Reference



360 Assembly[edit]

Translation of: AWK
*        count the coins           04/09/2015
COINS CSECT
USING COINS,R12
LR R12,R15
L R8,AMOUNT npenny=amount
L R4,AMOUNT
SRDA R4,32
D R4,=F'5'
LR R9,R5 nnickle=amount/5
L R4,AMOUNT
SRDA R4,32
D R4,=F'10'
LR R10,R5 ndime=amount/10
L R4,AMOUNT
SRDA R4,32
D R4,=F'25'
LR R11,R5 nquarter=amount/25
SR R1,R1 count=0
SR R4,R4 p=0
LOOPP CR R4,R8 do p=0 to npenny
BH ELOOPP
SR R5,R5 n=0
LOOPN CR R5,R9 do n=0 to nnickle
BH ELOOPN
SR R6,R6
LOOPD CR R6,R10 do d=0 to ndime
BH ELOOPD
SR R7,R7 q=0
LOOPQ CR R7,R11 do q=0 to nquarter
BH ELOOPQ
LR R3,R5 n
MH R3,=H'5'
LR R2,R4 p
AR R2,R3
LR R3,R6 d
MH R3,=H'10'
AR R2,R3
LR R3,R7 q
MH R3,=H'25'
AR R2,R3 s=p+n*5+d*10+q*25
C R2,=F'100' if s=100
BNE NOTOK
LA R1,1(R1) count=count+1
NOTOK LA R7,1(R7) q=q+1
B LOOPQ
ELOOPQ LA R6,1(R6) d=d+1
B LOOPD
ELOOPD LA R5,1(R5) n=n+1
B LOOPN
ELOOPN LA R4,1(R4) p=p+1
B LOOPP
ELOOPP XDECO R1,PG+0 edit count
XPRNT PG,12 print count
XR R15,R15
BR R14
AMOUNT DC F'100' start value in cents
PG DS CL12
YREGS
END COINS
Output:
         242

Ada[edit]

Works with: gnat/gcc
with Ada.Text_IO;
 
procedure Count_The_Coins is
 
type Counter_Type is range 0 .. 2**63-1; -- works with gnat
type Coin_List is array(Positive range <>) of Positive;
 
function Count(Goal: Natural; Coins: Coin_List) return Counter_Type is
Cnt: array(0 .. Goal) of Counter_Type := (0 => 1, others => 0);
-- 0 => we already know one way to choose (no) coins that sum up to zero
-- 1 .. Goal => we do not (yet) other ways to choose coins
begin
for C in Coins'Range loop
for Amount in 1 .. Cnt'Last loop
if Coins(C) <= Amount then
Cnt(Amount) := Cnt(Amount) + Cnt(Amount-Coins(C));
-- Amount-Coins(C) plus Coins(C) sums up to Amount;
end if;
end loop;
end loop;
return Cnt(Goal);
end Count;
 
procedure Print(C: Counter_Type) is
begin
Ada.Text_IO.Put_Line(Counter_Type'Image(C));
end Print;
 
begin
Print(Count( 1_00, (25, 10, 5, 1)));
Print(Count(1000_00, (100, 50, 25, 10, 5, 1)));
end Count_The_Coins;
Output:
 242
 13398445413854501

ALGOL 68[edit]

Works with: ALGOL 68G version Any - tested with release 2.4.1
Translation of: Haskell

This corresponds to a "naive" Haskell version; to do the larger problem will require a better approach.

 
#
Rosetta Code "Count the coins"
This is a direct translation of a Haskell version, using an array rather than
a list. LWB, UPB, and array slicing makes the mapping very simple:
 
LWB > UPB <=> []
LWB = UPB <=> [x]
a[LWB a] <=> head xs
a[LWB a + 1:] <=> tail xs
#

 
BEGIN
PROC ways to make change = ([] INT denoms, INT amount) INT :
BEGIN
IF amount = 0 THEN
1
ELIF LWB denoms > UPB denoms THEN
0
ELIF LWB denoms = UPB denoms THEN
(amount MOD denoms[LWB denoms] = 0 | 1 | 0)
ELSE
INT sum := 0;
FOR i FROM 0 BY denoms[LWB denoms] TO amount DO
sum +:= ways to make change(denoms[LWB denoms + 1:], amount - i)
OD;
sum
FI
END;
[] INT denoms = (25, 10, 5, 1);
print((ways to make change(denoms, 100), newline))
END
 
Output:
       +242

AutoHotkey[edit]

Translation of: Go
Works with: AutoHotkey_L
countChange(amount){
return cc(amount, 4)
}
 
cc(amount, kindsOfCoins){
if ( amount == 0 )
return 1
if ( amount < 0 ) || ( kindsOfCoins == 0 )
return 0
return cc(amount, kindsOfCoins-1)
+ cc(amount - firstDenomination(kindsOfCoins), kindsOfCoins)
}
 
firstDenomination(kindsOfCoins){
return [1, 5, 10, 25][kindsOfCoins]
}
MsgBox % countChange(100)

AWK[edit]

Iterative implementation, derived from Run BASIC:

#!/usr/bin/awk -f
 
BEGIN {
print cc(100)
exit
}
 
function cc(amount, coins, numPennies, numNickles, numQuarters, p, n, d, q, s, count) {
numPennies = amount
numNickles = int(amount / 5)
numDimes = int(amount / 10)
numQuarters = int(amount / 25)
 
count = 0
for (p = 0; p <= numPennies; p++) {
for (n = 0; n <= numNickles; n++) {
for (d = 0; d <= numDimes; d++) {
for (q = 0; q <= numQuarters; q++) {
s = p + n * 5 + d * 10 + q * 25;
if (s == 100) count++;
}
}
}
}
return count;
}
 

Run time:

time ./change-itr.awk
242

real	0m0.065s
user	0m0.063s
sys	0m0.002s

Recursive implementation (derived from Scheme example):

#!/usr/bin/awk -f
 
BEGIN {
COINSEP = ", "
coins = 1 COINSEP 5 COINSEP 10 COINSEP 25
print cc(100, coins)
exit
}
 
function cc(amt, coins) {
if (length(coins) == 0) return 0
if (amt < 0) return 0
if (amt == 0) return 1
return cc(amt, tail(coins)) + cc(amt - head(coins), coins)
}
 
function tail(coins, koins, s, c) {
split(coins, koins, COINSEP)
s = ""
for (c = 2; c <= length(koins); c++) s = s (s == "" ? "" : COINSEP) koins[c]
return s;
}
 
function head(coins, koins) {
split(coins, koins, COINSEP)
return koins[1]
}
 

Run time:

time ./change-rec.awk 
242

real	0m0.081s 
user	0m0.079s
sys	0m0.002s

While the recursive version is slower for small amounts, about 2 bucks it gets faster than the iterative version, at least until is segfaults from exhausting the stack.

BBC BASIC[edit]

Non-recursive solution:

      DIM uscoins%(3)
uscoins%() = 1, 5, 10, 25
PRINT FNchange(100, uscoins%()) " ways of making $1"
PRINT FNchange(1000, uscoins%()) " ways of making $10"
 
DIM ukcoins%(7)
ukcoins%() = 1, 2, 5, 10, 20, 50, 100, 200
PRINT FNchange(100, ukcoins%()) " ways of making £1"
PRINT FNchange(1000, ukcoins%()) " ways of making £10"
END
 
DEF FNchange(sum%, coins%())
LOCAL C%, D%, I%, N%, P%, Q%, S%, table()
C% = 0
N% = DIM(coins%(),1) + 1
FOR I% = 0 TO N% - 1
D% = coins%(I%)
IF D% <= sum% IF D% >= C% C% = D% + 1
NEXT
C% *= N%
DIM table(C%-1)
FOR I% = 0 TO N%-1 : table(I%) = 1 : NEXT
 
P% = N%
FOR S% = 1 TO sum%
FOR I% = 0 TO N% - 1
IF I% = 0 IF P% >= C% P% = 0
IF coins%(I%) <= S% THEN
Q% = P% - coins%(I%) * N%
IF Q% >= 0 table(P%) = table(Q%) ELSE table(P%) = table(Q% + C%)
ENDIF
IF I% table(P%) += table(P% - 1)
P% += 1
NEXT
NEXT
= table(P%-1)
 

Output (BBC BASIC does not have large enough integers for the optional task):

       242 ways of making $1
    142511 ways of making $10
      4563 ways of making £1
 321335886 ways of making £10

C[edit]

Using some crude 128-bit integer type.

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
 
// ad hoc 128 bit integer type; faster than using GMP because of low
// overhead
typedef struct { uint64_t x[2]; } i128;
 
// display in decimal
void show(i128 v) {
uint32_t x[4] = {v.x[0], v.x[0] >> 32, v.x[1], v.x[1] >> 32};
int i, j = 0, len = 4;
char buf[100];
do {
uint64_t c = 0;
for (i = len; i--; ) {
c = (c << 32) + x[i];
x[i] = c / 10, c %= 10;
}
 
buf[j++] = c + '0';
for (len = 4; !x[len - 1]; len--);
} while (len);
 
while (j--) putchar(buf[j]);
putchar('\n');
}
 
i128 count(int sum, int *coins)
{
int n, i, k;
for (n = 0; coins[n]; n++);
 
i128 **v = malloc(sizeof(int*) * n);
int *idx = malloc(sizeof(int) * n);
 
for (i = 0; i < n; i++) {
idx[i] = coins[i];
// each v[i] is a cyclic buffer
v[i] = calloc(sizeof(i128), coins[i]);
}
 
v[0][coins[0] - 1] = (i128) {{1, 0}};
 
for (k = 0; k <= sum; k++) {
for (i = 0; i < n; i++)
if (!idx[i]--) idx[i] = coins[i] - 1;
 
i128 c = v[0][ idx[0] ];
 
for (i = 1; i < n; i++) {
i128 *p = v[i] + idx[i];
 
// 128 bit addition
p->x[0] += c.x[0];
p->x[1] += c.x[1];
if (p->x[0] < c.x[0]) // carry
p->x[1] ++;
c = *p;
}
}
 
i128 r = v[n - 1][idx[n-1]];
 
for (i = 0; i < n; i++) free(v[i]);
free(v);
free(idx);
 
return r;
}
 
// simple recursive method; slow
int count2(int sum, int *coins)
{
if (!*coins || sum < 0) return 0;
if (!sum) return 1;
return count2(sum - *coins, coins) + count2(sum, coins + 1);
}
 
int main(void)
{
int us_coins[] = { 100, 50, 25, 10, 5, 1, 0 };
int eu_coins[] = { 200, 100, 50, 20, 10, 5, 2, 1, 0 };
 
show(count( 100, us_coins + 2));
show(count( 1000, us_coins));
 
show(count( 1000 * 100, us_coins));
show(count( 10000 * 100, us_coins));
show(count(100000 * 100, us_coins));
 
putchar('\n');
 
show(count( 1 * 100, eu_coins));
show(count( 1000 * 100, eu_coins));
show(count( 10000 * 100, eu_coins));
show(count(100000 * 100, eu_coins));
 
return 0;
}
output (only the first two lines are required by task):
242
13398445413854501
1333983445341383545001
133339833445334138335450001
 
4563
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

C#[edit]

 
// Adapted from http://www.geeksforgeeks.org/dynamic-programming-set-7-coin-change/
class Program
{
static long Count(int[] C, int m, int n)
{
var table = new long[n + 1];
table[0] = 1;
for (int i = 0; i < m; i++)
for (int j = C[i]; j <= n; j++)
table[j] += table[j - C[i]];
return table[n];
}
static void Main(string[] args)
{
var C = new int[] { 1, 5, 10, 25 };
int m = C.Length;
int n = 100;
Console.WriteLine(Count(C, m, n)); //242
Console.ReadLine();
}
}
 

Clojure[edit]

(def denomination-kind [1 5 10 25])
 
(defn- cc [amount denominations]
(cond (= amount 0) 1
(or (< amount 0) (empty? denominations)) 0
:else (+ (cc amount (rest denominations))
(cc (- amount (first denominations)) denominations))))
 
(defn count-change
"Calculates the number of times you can give change with the given denominations."
[amount denominations]
(cc amount denominations))
 
(count-change 15 denomination-kind) ; = 6

Coco[edit]

Translation of: Python
changes = (amount, coins) ->
ways = [1].concat [0] * amount
for coin of coins
for j from coin to amount
ways[j] += ways[j - coin]
ways[amount]
 
console.log changes 100, [1 5 10 25]

Common Lisp[edit]

Recursive Version With Cache[edit]

(defun count-change (amount coins
&optional
(length (1- (length coins)))
(cache (make-array (list (1+ amount) (length coins))
:initial-element nil)))
(cond ((< length 0) 0)
((< amount 0) 0)
((= amount 0) 1)
(t (or (aref cache amount length)
(setf (aref cache amount length)
(+ (count-change (- amount (first coins)) coins length cache)
(count-change amount (rest coins) (1- length) cache)))))))
 
; (compile 'count-change) ; for CLISP
 
(print (count-change 100 '(25 10 5 1))) ; = 242
(print (count-change 100000 '(100 50 25 10 5 1))) ; = 13398445413854501
(terpri)

Iterative Version[edit]

(defun count-change (amount coins &aux (ways (make-array (1+ amount) :initial-element 0)))
(setf (aref ways 0) 1)
(loop for coin in coins do
(loop for j from coin upto amount
do (incf (aref ways j) (aref ways (- j coin)))))
(aref ways amount))

D[edit]

Basic Version[edit]

Translation of: Go
import std.stdio, std.bigint;
 
auto changes(int amount, int[] coins) {
auto ways = new BigInt[amount + 1];
ways[0] = 1;
foreach (coin; coins)
foreach (j; coin .. amount + 1)
ways[j] += ways[j - coin];
return ways[$ - 1];
}
 
void main() {
changes( 1_00, [25, 10, 5, 1]).writeln;
changes(1000_00, [100, 50, 25, 10, 5, 1]).writeln;
}
Output:
242
13398445413854501

Safe Ulong Version[edit]

This version is very similar to the precedent, but it uses a faster ulong type, and performs a checked sum to detect overflows at run-time.

import std.stdio, core.checkedint;
 
auto changes(int amount, int[] coins, ref bool overflow) {
auto ways = new ulong[amount + 1];
ways[0] = 1;
foreach (coin; coins)
foreach (j; coin .. amount + 1)
ways[j] = ways[j].addu(ways[j - coin], overflow);
return ways[amount];
}
 
void main() {
bool overflow = false;
changes( 1_00, [25, 10, 5, 1], overflow).writeln;
if (overflow)
"Overflow".puts;
overflow = false;
changes( 1000_00, [100, 50, 25, 10, 5, 1], overflow).writeln;
if (overflow)
"Overflow".puts;
}

The output is the same.

Faster Version[edit]

Translation of: C
import std.stdio, std.bigint;
 
BigInt countChanges(in int amount, in int[] coins) pure /*nothrow*/ {
immutable n = coins.length;
int cycle;
foreach (immutable c; coins)
if (c <= amount && c >= cycle)
cycle = c + 1;
cycle *= n;
auto table = new BigInt[cycle];
table[0 .. n] = 1.BigInt;
 
int pos = n;
foreach (immutable s; 1 .. amount + 1) {
foreach (immutable i; 0 .. n) {
if (i == 0 && pos >= cycle)
pos = 0;
if (coins[i] <= s) {
immutable int q = pos - (coins[i] * n);
table[pos] = (q >= 0) ? table[q] : table[q + cycle];
}
if (i)
table[pos] += table[pos - 1];
pos++;
}
}
 
return table[pos - 1];
}
 
void main() {
immutable usCoins = [100, 50, 25, 10, 5, 1];
immutable euCoins = [200, 100, 50, 20, 10, 5, 2, 1];
 
foreach (immutable coins; [usCoins, euCoins]) {
countChanges( 1_00, coins[2 .. $]).writeln;
countChanges( 1000_00, coins).writeln;
countChanges( 10000_00, coins).writeln;
countChanges(100000_00, coins).writeln;
writeln;
}
}
Output:
242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

128-bit Version[edit]

A much faster version that mixes high-level and low-level style programming. This version uses basic 128-bit unsigned integers, like the C version. The output is the same as the second D version.

Translation of: C
import std.stdio, std.bigint, std.algorithm, std.conv, std.functional;
 
struct Ucent { /// Simplified 128-bit integer (like ucent).
ulong hi, lo;
static immutable one = Ucent(0, 1);
 
void opOpAssign(string op="+")(in ref Ucent y) pure nothrow @nogc @safe {
this.hi += y.hi;
if (this.lo >= ~y.lo)
this.hi++;
this.lo += y.lo;
}
 
string toString() const /*pure nothrow @safe*/ {
return text((this.hi.BigInt << 64) + this.lo);
}
}
 
Ucent countChanges(in int amount, in int[] coins) pure nothrow {
immutable n = coins.length;
 
// Points to a cyclic buffer of length coins[i]
auto p = new Ucent*[n];
auto q = new Ucent*[n]; // iterates it.
auto buf = new Ucent[coins.sum];
 
p[0] = buf.ptr;
foreach (immutable i; 0 .. n) {
if (i)
p[i] = coins[i - 1] + p[i - 1];
*p[i] = Ucent.one;
q[i] = p[i];
}
 
Ucent prev;
foreach (immutable j; 1 .. amount + 1)
foreach (immutable i; 0 .. n) {
q[i]--;
if (q[i] < p[i])
q[i] = p[i] + coins[i] - 1;
if (i)
*q[i] += prev;
prev = *q[i];
}
 
return prev;
}
 
void main() {
immutable usCoins = [100, 50, 25, 10, 5, 1];
immutable euCoins = [200, 100, 50, 20, 10, 5, 2, 1];
 
foreach (immutable coins; [usCoins, euCoins]) {
countChanges( 1_00, coins[2 .. $]).writeln;
countChanges( 1000_00, coins).writeln;
countChanges( 10000_00, coins).writeln;
countChanges(100000_00, coins).writeln;
writeln;
}
}

Printing Version[edit]

This version prints all the solutions (so it can be used on the smaller input):

import std.stdio, std.conv, std.string, std.algorithm, std.range;
 
void printChange(in uint tot, in uint[] coins)
in {
assert(coins.isSorted);
} body {
auto freqs = new uint[coins.length];
 
void inner(in uint curTot, in size_t start) {
if (curTot == tot)
return writefln("%-(%s %)",
zip(coins, freqs)
.filter!(cf => cf[1] != 0)
.map!(cf => format("%u:%u", cf[])));
 
foreach (immutable i; start .. coins.length) {
immutable ci = coins[i];
for (auto v = (freqs[i] + 1) * ci; v <= tot; v += ci)
if (curTot + v <= tot) {
freqs[i] += v / ci;
inner(curTot + v, i + 1);
freqs[i] -= v / ci;
}
}
}
 
inner(0, 0);
}
 
void main() {
printChange(1_00, [1, 5, 10, 25]);
}
Output:
1:5 5:1 10:4 25:2
1:5 5:1 10:9
1:5 5:2 10:1 25:3
1:5 5:2 10:6 25:1
1:5 5:3 10:3 25:2
1:5 5:3 10:8
1:5 5:4 10:5 25:1
1:5 5:4 25:3
1:5 5:5 10:2 25:2
1:5 5:5 10:7
1:5 5:6 10:4 25:1
1:5 5:7 10:1 25:2
...
5:11 10:2 25:1
5:12 10:4
5:13 10:1 25:1
5:14 10:3
5:15 25:1
5:16 10:2
5:18 10:1
5:20
10:5 25:2
10:10
25:4

EchoLisp[edit]

Recursive solution using memoization, adapted from CommonLisp and Racket.

 
(lib 'compile) ;; for (compile)
(lib 'bigint) ;; integer results > 32 bits
(lib 'hash) ;; hash table
 
;; h-table
(define Hcoins (make-hash))
 
;; the function to memoize
(define (sumways cents coins)
(+ (ways cents (cdr coins)) (ways (- cents (car coins)) coins)))
 
;; accelerator : ways (cents, coins) = ways ((cents - cents % 5) , coins)
(define (ways cents coins)
(cond ((null? coins) 0)
((negative? cents) 0)
((zero? cents) 1)
((eq? coins c-1) 1) ;; if coins = (1) --> 1
(else (hash-ref! Hcoins (list (- cents (modulo cents 5)) coins) sumways))))
 
(compile 'ways) ;; speed-up things
 
Output:
 
(define change '(25 10 5 1))
(define c-1 (list-tail change -1)) ;; pointer to (1)
(ways 100 change)
242
 
(define change '(100 50 25 10 5 1))
(define c-1 (list-tail change -1))
(for ((i (in-range 0 200001 20000)))
(writeln i (time (ways i change)) (hash-count Hcoins)))
 
 
;; iterate cents = 20000, 40000, ..
;; cents ((time (msec) number-of-ways) number-of-entries-in-h-table
 
20000 (350 4371565890901) 9398
40000 (245 138204514221801) 18798
60000 (230 1045248220992701) 28198
80000 (255 4395748062203601) 37598
100000 (234 13398445413854501) 46998
120000 (230 33312577651945401) 56398
140000 (292 71959878152476301) 65798
160000 (736 140236576291447201) 75198
180000 (237 252625397444858101) 84598
200000 (240 427707562988709001) 93998
 
;; One can see that the time is linear, and the h-table size reasonably small
 
change
(100 50 25 10 5 1)
(ways 100000 change)
13398445413854501
 
 


Elixir[edit]

Recursive Dynamic Programming solution in Elixir

defmodule Coins do
def find(coins,lim) do
vals = Enum.into(0..lim,Map.new,&{&1,0}) |> Map.put(0,1)
count(coins,lim,vals)
|> Map.values
|> Enum.max
|> IO.inspect
end
 
defp count([],_,vals), do: vals
defp count([coin|coins],lim,vals) do
count(coins,lim,ways(coin,coin,lim,vals))
end
 
defp ways(num,_coin,lim,vals) when num > lim, do: vals
defp ways(num, coin,lim,vals) do
ways(num+1,coin,lim,ad(coin,num,vals))
end
 
defp ad(a,b,c), do: Map.put(c,b,c[b]+c[b-a])
end
 
Coins.find([1,5,10,25],100)
Coins.find([1,5,10,25,50,100],100_000)
Output:
242
13398445413854501

Erlang[edit]

 
-module(coins).
-compile(export_all).
 
count(Amount, Coins) ->
{N,_C} = count(Amount, Coins, dict:new()),
N.
 
count(0,_,Cache) ->
{1,Cache};
count(N,_,Cache) when N < 0 ->
{0,Cache};
count(_N,[],Cache) ->
{0,Cache};
count(N,[C|Cs]=Coins,Cache) ->
case dict:is_key({N,length(Coins)},Cache) of
true ->
{dict:fetch({N,length(Coins)},Cache), Cache};
false ->
{N1,C1} = count(N-C,Coins,Cache),
{N2,C2} = count(N,Cs,C1),
{N1+N2,dict:store({N,length(Coins)},N1+N2,C2)}
end.
 
print(Amount, Coins) ->
io:format("~b ways to make change for ~b cents with ~p coins~n",[count(Amount,Coins),Amount,Coins]).
 
test() ->
A1 = 100, C1 = [25,10,5,1],
print(A1,C1),
A2 = 100000, C2 = [100, 50, 25, 10, 5, 1],
print(A2,C2).
 
Output:
42> coins:test().
242 ways to make change for 100 cents with [25,10,5,1] coins
13398445413854501 ways to make change for 100000 cents with [100,50,25,10,5,1] coins
ok

F#[edit]

Translation of: OCaml

Forward iteration, which can also be seen in Scala.

let changes amount coins =
let ways = Array.zeroCreate (amount + 1)
ways.[0] <- 1L
List.iter (fun coin ->
for j = coin to amount do ways.[j] <- ways.[j] + ways.[j - coin]
) coins
ways.[amount]
 
[<EntryPoint>]
let main argv =
printfn "%d" (changes 100 [25; 10; 5; 1]);
printfn "%d" (changes 100000 [100; 50; 25; 10; 5; 1]);
0
Output:
242
13398445413854501

Factor[edit]

USING: combinators kernel locals math math.ranges sequences sets sorting ;
IN: rosetta.coins
 
<PRIVATE
! recursive-count uses memoization and local variables.
! coins must be a sequence.
MEMO:: recursive-count ( cents coins -- ways )
coins length :> types
{
 ! End condition: 1 way to make 0 cents.
{ [ cents zero? ] [ 1 ] }
 ! End condition: 0 ways to make money without any coins.
{ [ types zero? ] [ 0 ] }
 ! Optimization: At most 1 way to use 1 type of coin.
{ [ types 1 number= ] [
cents coins first mod zero? [ 1 ] [ 0 ] if
] }
 ! Find all ways to use the first type of coin.
[
 ! f = first type, r = other types of coins.
coins unclip-slice :> f :> r
 ! Loop for 0, f, 2*f, 3*f, ..., cents.
0 cents f <range> [
 ! Recursively count how many ways to make remaining cents
 ! with other types of coins.
cents swap - r recursive-count
] [ + ] map-reduce  ! Sum the counts.
]
} cond ;
PRIVATE>
 
! How many ways can we make the given amount of cents
! with the given set of coins?
: make-change ( cents coins -- ways )
members [ ] inv-sort-with  ! Sort coins in descending order.
recursive-count ;

From the listener:

USE: rosetta.coins
( scratchpad ) 100 { 25 10 5 1 } make-change .
242
( scratchpad ) 100000 { 100 50 25 10 5 1 } make-change .
13398445413854501

This algorithm is slow. A test machine needed 1 minute to run 100000 { 100 50 25 10 5 1 } make-change . and get 13398445413854501. The same machine needed less than 1 second to run the Common Lisp (SBCL), Ruby (MRI) or Tcl (tclsh) programs and get the same answer.

Forth[edit]

\ counting change (SICP section 1.2.2)
 
: table create does> swap cells + @ ;
table coin-value 0 , 1 , 5 , 10 , 25 , 50 ,
 
: count-change ( total coin -- n )
over 0= if
2drop 1
else over 0< over 0= or if
2drop 0
else
2dup coin-value - over recurse
>r 1- recurse r> +
then then ;
 
100 5 count-change .

FutureBasic[edit]

 
include "ConsoleWindow"
 
dim as long penny, nickel, dime, quarter , count
 
penny = 1 : nickel = 1
dime = 1  : quarter = 1
 
for penny = 0 to 100
for nickel = 0 to 20
for dime = 0 to 10
for quarter = 0 to 4
if penny + nickel * 5 + dime * 10 + quarter * 25 == 100
print penny; " pennies "; nickel;" nickels "; dime; " dimes "; quarter; " quarters"
count++
end if
next quarter
next dime
next nickel
next penny
print count;" ways to make a dollar"
 
 

Output:

0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
0 pennies 1 nickels 2 dimes 3 quarters
......
65 pennies 5 nickels 1 dimes 0 quarters
65 pennies 7 nickels 0 dimes 0 quarters
70 pennies 0 nickels 3 dimes 0 quarters
70 pennies 1 nickels 0 dimes 1 quarters

242 ways to make a dollar


Go[edit]

A translation of the Lisp code referenced by the task description:

package main
 
import "fmt"
 
func main() {
amount := 100
fmt.Println("amount, ways to make change:", amount, countChange(amount))
}
 
func countChange(amount int) int64 {
return cc(amount, 4)
}
 
func cc(amount, kindsOfCoins int) int64 {
switch {
case amount == 0:
return 1
case amount < 0 || kindsOfCoins == 0:
return 0
}
return cc(amount, kindsOfCoins-1) +
cc(amount - firstDenomination(kindsOfCoins), kindsOfCoins)
}
 
func firstDenomination(kindsOfCoins int) int {
switch kindsOfCoins {
case 1:
return 1
case 2:
return 5
case 3:
return 10
case 4:
return 25
}
panic(kindsOfCoins)
}

Output:

amount, ways to make change: 100 242

Alternative algorithm, practical for the optional task.

package main
 
import "fmt"
 
func main() {
amount := 1000 * 100
fmt.Println("amount, ways to make change:", amount, countChange(amount))
}
 
func countChange(amount int) int64 {
ways := make([]int64, amount+1)
ways[0] = 1
for _, coin := range []int{100, 50, 25, 10, 5, 1} {
for j := coin; j <= amount; j++ {
ways[j] += ways[j-coin]
}
}
return ways[amount]
}

Output:

amount, ways to make change: 100000 13398445413854501

Groovy[edit]

Translation of: Go

Intuitive Recursive Solution:

def ccR
ccR = { BigInteger tot, List<BigInteger> coins ->
BigInteger n = coins.size()
switch ([tot:tot, coins:coins]) {
case { it.tot == 0 } :
return 1g
case { it.tot < 0 || coins == [] } :
return 0g
default:
return ccR(tot, coins[1..<n]) +
ccR(tot - coins[0], coins)
}
}

Fast Iterative Solution:

def ccI = { BigInteger tot, List<BigInteger> coins ->
List<BigInteger> ways = [0g] * (tot+1)
ways[0] = 1g
coins.each { BigInteger coin ->
(coin..tot).each { j ->
ways[j] += ways[j-coin]
}
}
ways[tot]
}

Test:

println '\nBase:'
[iterative: ccI, recursive: ccR].each { label, cc ->
print "${label} "
def start = System.currentTimeMillis()
def ways = cc(100g, [25g, 10g, 5g, 1g])
def elapsed = System.currentTimeMillis() - start
println ("answer: ${ways} elapsed: ${elapsed}ms")
}
 
print '\nExtra Credit:\niterative '
def start = System.currentTimeMillis()
def ways = ccI(1000g * 100, [100g, 50g, 25g, 10g, 5g, 1g])
def elapsed = System.currentTimeMillis() - start
println ("answer: ${ways} elapsed: ${elapsed}ms")

Output:

Base:
iterative answer: 242   elapsed: 5ms
recursive answer: 242   elapsed: 220ms

Extra Credit:
iterative answer: 13398445413854501   elapsed: 1077ms

Haskell[edit]

Naive implementation:

count 0 _ = 1
count _ [] = 0
count x (c:coins) = sum [ count (x - (n * c)) coins | n <- [0..(quot x c)] ]
 
main = print (count 100 [1, 5, 10, 25])

Much faster, probably harder to read, is to update results from bottom up:

count = foldr addCoin (1:repeat 0)
where addCoin c oldlist = newlist
where newlist = (take c oldlist) ++ zipWith (+) newlist (drop c oldlist)
 
main = do
print (count [25,10,5,1] !! 100)
print (count [100,50,25,10,5,1] !! 100000)

Icon and Unicon[edit]

procedure main()
 
US_coins := [1, 5, 10, 25]
US_allcoins := [1,5,10,25,50,100]
EU_coins := [1, 2, 5, 10, 20, 50, 100, 200]
CDN_coins := [1,5,10,25,100,200]
CDN_allcoins := [1,5,10,25,50,100,200]
 
every trans := ![ [15,US_coins],
[100,US_coins],
[1000*100,US_allcoins]
] do
printf("There are %i ways to count change for %i using %s coins.\n",CountCoins!trans,trans[1],ShowList(trans[2]))
end
 
procedure ShowList(L) # helper list to string
every (s := "[ ") ||:= !L || " "
return s || "]"
end

This is a naive implementation and very slow.

This example is in need of improvement:
Needs a better algorithm.
procedure CountCoins(amt,coins)  # very slow, recurse by coin value
local count
static S
 
if type(coins) == "list" then {
S := sort(set(coins))
if *S < 1 then runerr(205,coins)
return CountCoins(amt)
}
else {
/coins := 1
if value := S[coins] then {
every (count := 0) +:= CountCoins(amt - (0 to amt by value), coins + 1)
return count
}
else
return (amt ~= 0) | 1
}
end

printf.icn provides formatting

Output:
There are 6 ways to count change for 15 using [ 1 5 10 25 ] coins.
There are 242 ways to count change for 100 using [ 1 5 10 25 ] coins.
^c

Another one:

 
# coin.icn
# usage: coin value
procedure count(coinlist, value)
if value = 0 then return 1
if value < 0 then return 0
if (*coinlist <= 0) & (value >= 1) then return 0
return count(coinlist[1:*coinlist], value) + count(coinlist, value - coinlist[*coinlist])
end
 
 
procedure main(params)
money := params[1]
coins := [1,5,10,25]
 
writes("Value of ", money, " can be changed by using a set of ")
every writes(coins[1 to *coins], " ")
write(" coins in ", count(coins, money), " different ways.")
end
 

Output:

Value of 15 can be changed by using a set of 1 5 10 25  coins in 6 different ways.
Value of 100 can be changed by using a set of 1 5 10 25  coins in 242 different ways.

J[edit]

In this draft intermediate results are a two column array. The first column is tallies -- the number of ways we have for reaching the total represented in the second column, which is unallocated value (which we will assume are pennies). We will have one row for each different in-range value which can be represented using only nickles (0, 5, 10, ... 95, 100).

merge=: ({:"1 (+/@:({."1),{:@{:)/. ])@;
count=: {.@] <@,. {:@] - [ * [ i.@>:@<.@%~ {:@]
init=: (1 ,. ,.)^:(0=#@$)
nsplits=: 0 { [: +/ [: (merge@:(count"1) init)/ }.@/:~@~.@,

This implementation special cases the handling of pennies and assumes that the lowest coin value in the argument is 1. If I needed additional performance, I would next special case the handling of nickles/penny combinations...

Thus:

   100 nsplits 1 5 10 25
242

And, on a 64 bit machine with sufficient memory:

   100000 nsplits 1 5 10 25 50 100
13398445413854501

Java[edit]

Translation of: D
Works with: Java version 1.5+
import java.util.Arrays;
import java.math.BigInteger;
 
class CountTheCoins {
private static BigInteger countChanges(int amount, int[] coins){
final int n = coins.length;
int cycle = 0;
for (int c : coins)
if (c <= amount && c >= cycle)
cycle = c + 1;
cycle *= n;
BigInteger[] table = new BigInteger[cycle];
Arrays.fill(table, 0, n, BigInteger.ONE);
Arrays.fill(table, n, cycle, BigInteger.ZERO);
 
int pos = n;
for (int s = 1; s <= amount; s++) {
for (int i = 0; i < n; i++) {
if (i == 0 && pos >= cycle)
pos = 0;
if (coins[i] <= s) {
final int q = pos - (coins[i] * n);
table[pos] = (q >= 0) ? table[q] : table[q + cycle];
}
if (i != 0)
table[pos] = table[pos].add(table[pos - 1]);
pos++;
}
}
 
return table[pos - 1];
}
 
public static void main(String[] args) {
final int[][] coinsUsEu = {{100, 50, 25, 10, 5, 1},
{200, 100, 50, 20, 10, 5, 2, 1}};
 
for (int[] coins : coinsUsEu) {
System.out.println(countChanges( 100,
Arrays.copyOfRange(coins, 2, coins.length)));
System.out.println(countChanges( 100000, coins));
System.out.println(countChanges( 1000000, coins));
System.out.println(countChanges(10000000, coins) + "\n");
}
}
}

Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

JavaScript[edit]

Iterative[edit]

Efficient iterative algorithm (cleverly calculates number of combinations without permuting them)

 
function countcoins(t, o) {
'use strict';
var targetsLength = t + 1;
var operandsLength = o.length;
t = [1];
 
for (var a = 0; a < operandsLength; a ++) {
for (var b = 1; b < targetsLength; b ++) {
 
// initialise undefined target
t[b] = t[b] ? t[b] : 0;
 
// accumulate target + operand ways
t[b] += (b < o[a]) ? 0 : t[b - o[a]];
}
}
 
return t[targetsLength - 1];
}
 
Output:

JavaScript hits integer limit for optional task

 
countcoins(100, [1,5,10,25]);
242
 

Recursive[edit]

Inefficient recursive algorithm (naively calculates number of combinations by actually permuting them)

 
function countcoins(t, o) {
'use strict';
var operandsLength = o.length;
var solutions = 0;
 
function permutate(a, x) {
 
// base case
if (a === t) {
solutions ++;
}
 
// recursive case
else if (a < t) {
for (var i = 0; i < operandsLength; i ++) {
if (i >= x) {
permutate(o[i] + a, i);
}
}
}
}
 
permutate(0, 0);
return solutions;
}
 
Output:

Too slow for optional task

 
countcoins(100, [1,5,10,25]);
242
 

Iterative again[edit]

Translation of: C#
var amount=100, coin=[1,5,10,25]
var t=[1]; for (t[amount]=0, a=1; a<amount; a++) t[a]=0 // initialise t[0..amount]=[1,0,...,0]
for (var i=0, e=coin.length; i<e; i++)
for (var ci=coin[i], a=ci; a<=amount; a++)
t[a] += t[a-ci]
document.write( t[amount] )
Output:
242

jq[edit]

Currently jq uses IEEE 754 64-bit numbers. Large integers are approximated by floats, and therefore the answer that the following program provides for the optional task is only correct for the first 15 digits.

# How many ways are there to make "target" cents, given a list of coin
# denominations as input.
# The strategy is to record at total[n] the number of ways to make n cents.
def countcoins(target):
. as $coin
| reduce range(0; length) as $a
( [1]; # there is 1 way to make 0 cents
reduce range(1; target + 1) as $b
(.; # total[]
if $b < $coin[$a] then .
else .[$b - $coin[$a]] as $count
| if $count == 0 then .
else .[$b] += $count
end
end ) )
| .[target] ;

Example:

[1,5,10,25] | countcoins(100)
Output:
242

Lasso[edit]

Inspired by the javascript iterative example for the same task

define cointcoins(
target::integer,
operands::array
) => {
 
local(
targetlength = #target + 1,
operandlength = #operands -> size,
output = staticarray_join(#targetlength,0),
outerloopcount
)
 
#output -> get(1) = 1
 
loop(#operandlength) => {
#outerloopcount = loop_count
loop(#targetlength) => {
 
if(loop_count >= #operands -> get(#outerloopcount) and loop_count - #operands -> get(#outerloopcount) > 0) => {
#output -> get(loop_count) += #output -> get(loop_count - #operands -> get(#outerloopcount))
}
}
}
 
return #output -> get(#targetlength)
}
 
cointcoins(100, array(1,5,10,25,))
'<br />'
cointcoins(100000, array(1, 5, 10, 25, 50, 100))

Output:

242
13398445413854501

Lua[edit]

Lua uses one-based indexes but table keys can be any value so you can define an element 0 just as easily as you can define an element "foo"...

function countSums (amount, values)
local t = {}
for i = 1, amount do t[i] = 0 end
t[0] = 1
for k, val in pairs(values) do
for i = val, amount do t[i] = t[i] + t[i - val] end
end
return t[amount]
end
 
print(countSums(100, {1, 5, 10, 25}))
print(countSums(100000, {1, 5, 10, 25, 50, 100}))
Output:
242
1.3398445413855e+16

Mathematica / Wolfram Language[edit]

Translation of: Go
CountCoins[amount_, coinlist_] := ( ways = ConstantArray[1, amount];
Do[For[j = coin, j <= amount, j++,
If[ j - coin == 0,
ways[[j]] ++,
ways[[j]] += ways[[j - coin]]
]]
, {coin, coinlist}];
ways[[amount]])

Example usage:

CountCoins[100, {25, 10, 5}]
-> 242

CountCoins[100000, {100, 50, 25, 10, 5}]
-> 13398445413854501

MATLAB / Octave[edit]

 
%% Count_The_Coins
clear;close all;clc;
tic
 
for i = 1:2 % 1st loop is main challenge 2nd loop is optional challenge
if (i == 1)
amount = 100; % Matlab indexes from 1 not 0, so we need to add 1 to our target value
amount = amount + 1;
coins = [1 5 10 25]; % Value of coins we can use
else
amount = 100*1000; % Matlab indexes from 1 not 0, so we need to add 1 to our target value
amount = amount + 1;
coins = [1 5 10 25 50 100]; % Value of coins we can use
end % End if
ways = zeros(1,amount); % Preallocating for speed
ways(1) = 1; % First solution is 1
 
% Solves from smallest sub problem to largest (bottom up approach of dynamic programming).
for j = 1:length(coins)
for K = coins(j)+1:amount
ways(K) = ways(K) + ways(K-coins(j));
end % End for
end % End for
if (i == 1)
fprintf(‘Main Challenge: %d \n', ways(amount));
else
fprintf(‘Bonus Challenge: %d \n', ways(amount));
end % End if
end % End for
toc
 

Example Output:

Main Challenge: 242

Bonus Challenge: 13398445413854501

Mercury[edit]

:- module coins.
:- interface.
:- import_module int, io.
:- type coin ---> quarter; dime; nickel; penny.
:- type purse ---> purse(int, int, int, int).
 
:- pred sum_to(int::in, purse::out) is nondet.
 
:- pred main(io::di, io::uo) is det.
:- implementation.
:- import_module solutions, list, string.
 
:- func value(coin) = int.
value(quarter) = 25.
value(dime) = 10.
value(nickel) = 5.
value(penny) = 1.
 
:- pred supply(coin::in, int::in, int::out) is multi.
supply(C, Target, N) :- upto(Target div value(C), N).
 
:- pred upto(int::in, int::out) is multi.
upto(N, R) :- ( nondet_int_in_range(0, N, R0) -> R = R0 ; R = 0 ).
 
sum_to(To, Purse) :-
Purse = purse(Q, D, N, P),
sum(Purse) = To,
supply(quarter, To, Q),
supply(dime, To, D),
supply(nickel, To, N),
supply(penny, To, P).
 
:- func sum(purse) = int.
sum(purse(Q, D, N, P)) =
value(quarter) * Q + value(dime) * D +
value(nickel) * N + value(penny) * P.
 
main(!IO) :-
solutions(sum_to(100), L),
show(L, !IO),
io.format("There are %d ways to make change for a dollar.\n",
[i(length(L))], !IO).
 
:- pred show(list(purse)::in, io::di, io::uo) is det.
show([], !IO).
show([P|T], !IO) :-
io.write(P, !IO), io.nl(!IO),
show(T, !IO).

Nim[edit]

Translation of: Python
proc changes(amount, coins): int =
var ways = @[1]
ways.setLen(amount+1)
for coin in coins:
for j in coin..amount:
ways[j] += ways[j-coin]
ways[amount]
 
echo changes(100, [1, 5, 10, 25])
echo changes(100000, [1, 5, 10, 25, 50, 100])

Output:

242
13398445413854501

OCaml[edit]

Translation of the D minimal version:

let changes amount coins =
let ways = Array.make (amount + 1) 0L in
ways.(0) <- 1L;
List.iter (fun coin ->
for j = coin to amount do
ways.(j) <- Int64.add ways.(j) ways.(j - coin)
done
) coins;
ways.(amount)
 
let () =
Printf.printf "%Ld\n" (changes 1_00 [25; 10; 5; 1]);
Printf.printf "%Ld\n" (changes 1000_00 [100; 50; 25; 10; 5; 1]);
;;

Output:

$ ocaml coins.ml 
242
13398445413854501

PARI/GP[edit]

coins(v)=prod(i=1,#v,1/(1-'x^v[i]));
ways(v,n)=polcoeff(coins(v)+O('x^(n+1)),n);
ways([1,5,10,25],100)
ways([1,5,10,25,50,100],100000)

Output:

%1 = 242
%2 = 13398445413854501

Perl[edit]

use 5.01;
use Memoize;
 
sub cc {
my $amount = shift;
return 0 if !@_ || $amount < 0;
return 1 if $amount == 0;
my $first = shift;
cc( $amount, @_ ) + cc( $amount - $first, $first, @_ );
}
memoize 'cc';
 
# Make recursive algorithm run faster by sorting coins descending by value:
sub cc_optimized {
my $amount = shift;
cc( $amount, sort { $b <=> $a } @_ );
}
 
say 'Ways to change $ 1 with common coins: ',
cc_optimized( 100, 1, 5, 10, 25 );
say 'Ways to change $ 1000 with addition of less common coins: ',
cc_optimized( 1000 * 100, 1, 5, 10, 25, 50, 100 );
 
Output:
Ways to change $ 1 with common coins: 242
Ways to change $ 1000 with addition of less common coins: 13398445413854501

Perl 6[edit]

Works with: rakudo version 2015.09
Translation of: Ruby

Recursive (cached)[edit]

sub ways-to-make-change($amount, @coins) {
my @cache = $[1 xx @coins];
 
multi ways($n where $n >= 0, @now [$coin,*@later]) {
@cache[$n][+@later] //= ways($n - $coin, @now) + ways($n, @later);
}
multi ways($,@) { 0 }
 
ways($amount, @coins.sort(-*).list); # sort descending
}
 
say ways-to-make-change 1_00, [1,5,10,25];
say ways-to-make-change 1000_00, [1,5,10,25,50,100];
Output:
242
13398445413854501

Iterative[edit]

sub ways-to-make-change-slowly(\n, @coins) {
my @table = [1 xx @coins], [0 xx @coins] xx n;
for 1..n X ^@coins -> (\i, \j) {
my \c = @coins[j];
@table[i][j] = [+]
@table[i - c][j ] // 0,
@table[i ][j - 1] // 0;
}
@table[*-1][*-1];
}
 
say ways-to-make-change-slowly 1_00, [1,5,10,25];
say ways-to-make-change-slowly 1000_00, [1,5,10,25,50,100];

PicoLisp[edit]

Translation of: C
(de coins (Sum Coins)
(let (Buf (mapcar '((N) (cons 1 (need (dec N) 0))) Coins) Prev)
(do Sum
(zero Prev)
(for L Buf
(inc (rot L) Prev)
(setq Prev (car L)) ) )
Prev ) )

Test:

(for Coins '((100 50 25 10 5 1) (200 100 50 20 10 5 2 1))
(println (coins 100 (cddr Coins)))
(println (coins (* 1000 100) Coins))
(println (coins (* 10000 100) Coins))
(println (coins (* 100000 100) Coins))
(prinl) )

Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

Python[edit]

Simple version[edit]

Translation of: Go
def changes(amount, coins):
ways = [0] * (amount + 1)
ways[0] = 1
for coin in coins:
for j in xrange(coin, amount + 1):
ways[j] += ways[j - coin]
return ways[amount]
 
print changes(100, [1, 5, 10, 25])
print changes(100000, [1, 5, 10, 25, 50, 100])

Output:

242
13398445413854501

Fast version[edit]

Translation of: C
try:
import psyco
psyco.full()
except ImportError:
pass
 
def count_changes(amount_cents, coins):
n = len(coins)
# max([]) instead of max() for Psyco
cycle = max([c+1 for c in coins if c <= amount_cents]) * n
table = [0] * cycle
for i in xrange(n):
table[i] = 1
 
pos = n
for s in xrange(1, amount_cents + 1):
for i in xrange(n):
if i == 0 and pos >= cycle:
pos = 0
if coins[i] <= s:
q = pos - coins[i] * n
table[pos]= table[q] if (q >= 0) else table[q + cycle]
if i:
table[pos] += table[pos - 1]
pos += 1
return table[pos - 1]
 
def main():
us_coins = [100, 50, 25, 10, 5, 1]
eu_coins = [200, 100, 50, 20, 10, 5, 2, 1]
 
for coins in (us_coins, eu_coins):
print count_changes( 100, coins[2:])
print count_changes( 100000, coins)
print count_changes( 1000000, coins)
print count_changes(10000000, coins), "\n"
 
main()

Output:

242
13398445413854501
1333983445341383545001
133339833445334138335450001

4562
10056050940818192726001
99341140660285639188927260001
992198221207406412424859964272600001

Racket[edit]

This is the basic recursive way:

#lang racket
(define (ways-to-make-change cents coins)
(cond ((null? coins) 0)
((negative? cents) 0)
((zero? cents) 1)
(else
(+ (ways-to-make-change cents (cdr coins))
(ways-to-make-change (- cents (car coins)) coins)))))
 
(ways-to-make-change 100 '(25 10 5 1)) ; -> 242
 

This works for the small numbers, but the optional task is just too slow with this solution, so with little change to the code we can use memoization:

#lang racket
 
(define memos (make-hash))
(define (ways-to-make-change cents coins)
(cond [(or (empty? coins) (negative? cents)) 0]
[(zero? cents) 1]
[else (define (answerer-for-new-arguments)
(+ (ways-to-make-change cents (rest coins))
(ways-to-make-change (- cents (first coins)) coins)))
(hash-ref! memos (cons cents coins) answerer-for-new-arguments)]))
 
(time (ways-to-make-change 100 '(25 10 5 1)))
(time (ways-to-make-change 100000 '(100 50 25 10 5 1)))
(time (ways-to-make-change 1000000 '(200 100 50 20 10 5 2 1)))
 
#| Times in milliseconds, and results:
 
cpu time: 1 real time: 1 gc time: 0
242
 
cpu time: 524 real time: 553 gc time: 163
13398445413854501
 
cpu time: 20223 real time: 20673 gc time: 10233
99341140660285639188927260001 |#

REXX[edit]

recursive[edit]

The recursive calls to the subroutine have been unrolled somewhat, this reduces the number of recursive calls substantially.

These REXX versions also support fractional cents (as in a   ½-cent   and   ¼-cent coins).   Any fractional coin can be
specified as a decimal fraction   (.5,    .25,   ···).

Support was included to allow specification of half-cent and quarter-cent coins as   1/2   and   1/4.

The amount can be specified in cents (as a number), or in dollars (as for instance,   $1000).

/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20 /*be able to handle large amounts of $.*/
parse arg N $ /*obtain optional arguments from the CL*/
if N='' | N="," then N=100 /*Not specified? Then Use $1 (≡100¢).*/
if $='' | $="," then $=1 5 10 25 /*Use penny/nickel/dime/quarter default*/
if left(N,1)=='$' then N=100*substr(N,2) /*the amount was specified in dollars.*/
coins=words($) /*the number of coins specified. */
NN=N; do j=1 for coins /*create a fast way of accessing specie*/
_=word($,j) /*define an array element for the coin.*/
if _=='1/2' then _=.5 /*an alternate spelling of a half-cent.*/
if _=='1/4' then _=.25 /* " " " " " quarter-¢.*/
$.j=_ /*assign the value to a particular coin*/
end /*j*/
_=n//100; cnt=' cents' /* [↓] is the amount in whole dollars?*/
if _=0 then do; NN='$' || (NN%100); cnt=; end /*show the amount in dollars, not cents*/
say 'with an amount of ' commas(NN)cnt", there are " commas( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: ' $
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: procedure; parse arg _; n=_'.9'; #=123456789; b=verify(n,#,"M")
e=verify(n,#'0',,verify(n,#"0.",'M'))-4
do j=e to b by -3; _=insert(',',_,j); end /*j*/; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $.; parse arg a,k /*this function is invoked recursively.*/
if a==0 then return 1 /*unroll for a special case of zero. */
if k==1 then return 1 /* " " " " " " unity. */
if k==2 then f=1 /*handle this special case of two. */
else f=MKchg(a, k-1) /*count, and then recurse the amount. */
if a==$.k then return f+1 /*handle this special case of A=a coin.*/
if a <$.k then return f /* " " " " " A<a coin.*/
return f+MKchg(a-$.k,k) /*use diminished amount ($) for change.*/

output   when using the default input:

with an amount of  $1,  there are  242
ways to make change with coins of the following denominations:  1 5 10 25

output   when using the following input:   $1   1/4   1/2   1   2   3   5   10   20   25   50   100

with an amount of  $1,  there are  29,034,171
ways to make change with coins of the following denominations:  1/4 1/2 1 2 3 5 10 20 25 50 100

with memoization[edit]

This REXX version is more than a couple of orders of magnitude faster than the 1st version when using larger amounts.

/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20 /*be able to handle large amounts of $.*/
parse arg N $ /*obtain optional arguments from the CL*/
if N='' | N="," then N=100 /*Not specified? Then Use $1 (≡100¢).*/
if $='' | $="," then $=1 5 10 25 /*Use penny/nickel/dime/quarter default*/
if left(N,1)=='$' then N=100*substr(N,2) /*the amount was specified in dollars.*/
coins=words($) /*the number of coins specified. */
!.=.; NN=N; do j=1 for coins /*create a fast way of accessing specie*/
_=word($,j);  ?=_ ' coin' /*define an array element for the coin.*/
if _=='1/2' then _=.5 /*an alternate spelling of a half-cent.*/
if _=='1/4' then _=.25 /* " " " " " quarter-¢.*/
$.j=_ /*assign the value to a particular coin*/
end /*j*/
_=n//100; cnt=' cents' /* [↓] is the amount in whole dollars?*/
if _=0 then do; NN='$' || (NN%100); cnt=; end /*show the amount in dollars, not cents*/
say 'with an amount of ' commas(NN)cnt", there are " commas( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: ' $
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: procedure; parse arg _; n=_'.9'; #=123456789; b=verify(n,#,"M")
e=verify(n,#'0',,verify(n,#"0.",'M'))-4
do j=e to b by -3; _=insert(',',_,j); end /*j*/; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $. !.; parse arg a,k /*function is recursive. */
if !.a.k\==. then return !.a.k /*found this A & K before? */
if a==0 then return 1 /*unroll for a special case*/
if k==1 then return 1 /* " " " " " */
if k==2 then f=1 /*handle this special case.*/
else f=MKchg(a, k-1) /*count, recurse the amount*/
if a==$.k then do; !.a.k=f+1; return !.a.k; end /*handle this special case.*/
if a <$.k then do; !.a.k=f  ; return f  ; end /* " " " " */
 !.a.k=f + MKchg(a-$.k, k); return !.a.k /*compute, define, return. */

output   when using the following input for the optional test case:   $1000   1   5   10   25   50   100

with an amount of  $1,000,  there are  13,398,445,413,854,501
ways to make change with coins of the following denominations:  1 5 10 25 50 100

with error checking[edit]

This REXX version is identical to the previous REXX version, but has error checking for the amount and the coins specified.

/*REXX program counts the number of ways to make change with coins from an given amount.*/
numeric digits 20 /*be able to handle large amounts of $.*/
parse arg N $ /*obtain optional arguments from the CL*/
if N='' | N="," then N=100 /*Not specified? Then Use $1 (≡100¢).*/
if $='' | $="," then $=1 5 10 25 /*Use penny/nickel/dime/quarter default*/
X=N /*save original for possible error msgs*/
if left(N,1)=='$' then do /*the amount has a leading dollar sign.*/
_=substr(N,2) /*the amount was specified in dollars.*/
if \isNum(_) then call ser "amount isn't numeric: " N
N=100*_ /*change amount (in $) ───► cents (¢).*/
end
max$=10**digits() /*the maximum amount this pgm can have.*/
if \isNum(N) then call ser X " amount isn't numeric."
if N=0 then call ser X " amount can't be zero."
if N<0 then call ser X " amount can't be negative."
if N>max$ then call ser X " amount can't be greater than " max$'.'
coins=words($);  !.=.; NN=N; p=0 /*#coins specified; coins; amount; prev*/
@.=0 /*verify a coin was only specified once*/
do j=1 for coins /*create a fast way of accessing specie*/
_=word($,j);  ?=_ ' coin' /*define an array element for the coin.*/
if _=='1/2' then _=.5 /*an alternate spelling of a half-cent.*/
if _=='1/4' then _=.25 /* " " " " " quarter-¢.*/
if \isNum(_) then call ser ? "coin value isn't numeric."
if _<0 then call ser ? "coin value can't be negative."
if _<=0 then call ser ? "coin value can't be zero."
if @._ then call ser ? "coin was already specified."
if _<p then call ser ? "coin must be greater than previous:" p
if _>N then call ser ? "coin must be less or equal to amount:" X
@._=1; p=_ /*signify coin was specified; set prev.*/
$.j=_ /*assign the value to a particular coin*/
end /*j*/
_=n//100; cnt=' cents' /* [↓] is the amount in whole dollars?*/
if _=0 then do; NN='$' || (NN%100); cnt=; end /*show the amount in dollars, not cents*/
say 'with an amount of ' commas(NN)cnt", there are " commas( MKchg(N, coins) )
say 'ways to make change with coins of the following denominations: ' $
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
isNum: return datatype(arg(1), 'N') /*return 1 if arg is numeric, 0 if not.*/
ser: say; say '***error***'; say; say arg(1); say; exit 13 /*error msg.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: procedure; parse arg _; n=_'.9'; #=123456789; b=verify(n,#,"M")
e=verify(n,#'0',,verify(n,#"0.",'M'))-4
do j=e to b by -3; _=insert(',',_,j); end /*j*/; return _
/*──────────────────────────────────────────────────────────────────────────────────────*/
MKchg: procedure expose $. !.; parse arg a,k /*function is recursive. */
if !.a.k\==. then return !.a.k /*found this A & K before? */
if a==0 then return 1 /*unroll for a special case*/
if k==1 then return 1 /* " " " " " */
if k==2 then f=1 /*handle this special case.*/
else f=MKchg(a, k-1) /*count, recurse the amount*/
if a==$.k then do; !.a.k=f+1; return !.a.k; end /*handle this special case.*/
if a <$.k then do; !.a.k=f  ; return f  ; end /* " " " " */
 !.a.k=f + MKchg(a-$.k, k); return !.a.k /*compute, define, return. */

output   is the same as the previous REXX versions.

Ruby[edit]

The algorithm also appears here

Recursive, with caching

def make_change(amount, coins)
@cache = Array.new(amount+1){|i| Array.new(coins.size, i.zero? ? 1 : nil)}
@coins = coins
do_count(amount, @coins.length - 1)
end
 
def do_count(n, m)
if n < 0 || m < 0
0
elsif @cache[n][m]
@cache[n][m]
else
@cache[n][m] = do_count(n-@coins[m], m) + do_count(n, m-1)
end
end
 
p make_change( 1_00, [1,5,10,25])
p make_change(1000_00, [1,5,10,25,50,100])

outputs

242
13398445413854501

Iterative

def make_change2(amount, coins)
n, m = amount, coins.size
table = Array.new(n+1){|i| Array.new(m, i.zero? ? 1 : nil)}
for i in 1..n
for j in 0...m
table[i][j] = (i<coins[j] ? 0 : table[i-coins[j]][j]) +
(j<1  ? 0 : table[i][j-1])
end
end
table[-1][-1]
end
 
p make_change2( 1_00, [1,5,10,25])
p make_change2(1000_00, [1,5,10,25,50,100])

outputs

242
13398445413854501


Run BASIC[edit]

for penny         = 0 to 100
for nickel = 0 to 20
for dime = 0 to 10
for quarter = 0 to 4
if penny + nickel * 5 + dime * 10 + quarter * 25 = 100 then
print penny;" pennies ";nickel;" nickels "; dime;" dimes ";quarter;" quarters"
count = count + 1
end if
next quarter
next dime
next nickel
next penny
print count;" ways to make a buck"
Output:
0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
0 pennies 1 nickels 2 dimes 3 quarters
......
65 pennies 5 nickels 1 dimes 0 quarters
65 pennies 7 nickels 0 dimes 0 quarters
70 pennies 0 nickels 3 dimes 0 quarters
70 pennies 1 nickels 0 dimes 1 quarters
.....
242 ways to make a buck

SAS[edit]

Generate the solutions using CLP solver in SAS/OR:

/* call OPTMODEL procedure in SAS/OR */
proc optmodel;
/* declare set and names of coins */
set COINS = {1,5,10,25};
str name {COINS} = ['penny','nickel','dime','quarter'];
 
/* declare variables and constraint */
var NumCoins {COINS} >= 0 integer;
con Dollar:
sum {i in COINS} i * NumCoins[i] = 100;
 
/* call CLP solver */
solve with CLP / findallsolns;
 
/* write solutions to SAS data set */
create data sols(drop=s) from [s]=(1.._NSOL_) {i in COINS} <col(name[i])=NumCoins[i].sol[s]>;
quit;
 
/* print all solutions */
proc print data=sols;
run;

Output:

Obs penny nickel dime quarter 
1 100 0 0 0 
2 95 1 0 0 
3 90 2 0 0 
4 85 3 0 0 
5 80 4 0 0 
...
238 5 2 1 3 
239 0 3 1 3 
240 5 0 2 3 
241 0 1 2 3 
242 0 0 0 4 

Scala[edit]

def countChange(amount: Int, coins:List[Int]) = {
val ways = Array.fill(amount + 1)(0)
ways(0) = 1
coins.foreach (coin =>
for (j<-coin to amount)
ways(j) = ways(j) + ways(j - coin)
)
ways(amount)
}
 
countChange (15, List(1, 5, 10, 25))
 

Output:

res0: Int = 6

Recursive implementation:

def count(target: Int, coins: List[Int]): Int = {
if (target == 0) 1
else if (coins.isEmpty || target < 0) 0
else count(target, coins.tail) + count(target - coins.head, coins)
}
 
 
count(100, List(25, 10, 5, 1))
 

Scheme[edit]

A simple recursive implementation:

(define ways-to-make-change
(lambda (x coins)
(cond
[(null? coins) 0]
[(< x 0) 0]
[(zero? x) 1]
[else (+ (ways-to-make-change x (cdr coins)) (ways-to-make-change (- x (car coins)) coins))])))
 
(ways-to-make-change 100)

Output:

242

Seed7[edit]

$ include "seed7_05.s7i";
include "bigint.s7i";
 
const func bigInteger: changeCount (in integer: amountCents, in array integer: coins) is func
result
var bigInteger: waysToChange is 0_;
local
var array bigInteger: t is 0 times 0_;
var integer: pos is 0;
var integer: s is 0;
var integer: i is 0;
begin
t := length(coins) times 1_ & (length(coins) * amountCents) times 0_;
pos := length(coins) + 1;
for s range 1 to amountCents do
if coins[1] <= s then
t[pos] := t[pos - (length(coins) * coins[1])];
end if;
incr(pos);
for i range 2 to length(coins) do
if coins[i] <= s then
t[pos] := t[pos - (length(coins) * coins[i])];
end if;
t[pos] +:= t[pos - 1];
incr(pos);
end for;
end for;
waysToChange := t[pos - 1];
end func;
 
const proc: main is func
local
const array integer: usCoins is [] (1, 5, 10, 25, 50, 100);
const array integer: euCoins is [] (1, 2, 5, 10, 20, 50, 100, 200);
begin
writeln(changeCount( 100, usCoins[.. 4]));
writeln(changeCount( 100000, usCoins));
writeln(changeCount(1000000, usCoins));
writeln(changeCount( 100000, euCoins));
writeln(changeCount(1000000, euCoins));
end func;

Output:

242
13398445413854501
1333983445341383545001
10056050940818192726001
99341140660285639188927260001

Sidef[edit]

Translation of: Perl
func cc(_)                { 0 }
func cc({ .is_neg }, *_) { 0 }
func cc({ .is_zero }, *_) { 1 }
 
func cc(amount, first, *rest) is cached {
cc(amount, rest...) + cc(amount - first, first, rest...);
}
 
func cc_optimized(amount, *rest) {
cc(amount, rest.sort_by{|v| -v }...);
}
 
var x = cc_optimized(100, 1, 5, 10, 25);
say "Ways to change $1 with common coins: #{x}";
 
var y = cc_optimized(1000 * 100, 1, 5, 10, 25, 50, 100);
say "Ways to change $1000 with addition of less common coins: #{y}";
Output:
Ways to change $1 with common coins: 242
Ways to change $1000 with addition of less common coins: 13398445413854501

Tcl[edit]

Translation of: Ruby
package require Tcl 8.5
 
proc makeChange {amount coins} {
set table [lrepeat [expr {$amount+1}] [lrepeat [llength $coins] {}]]
lset table 0 [lrepeat [llength $coins] 1]
for {set i 1} {$i <= $amount} {incr i} {
for {set j 0} {$j < [llength $coins]} {incr j} {
set k [expr {$i - [lindex $coins $j]}]
lset table $i $j [expr {
($k < 0 ? 0 : [lindex $table $k $j]) +
($j < 1 ? 0 : [lindex $table $i [expr {$j-1}]])
}]
}
}
return [lindex $table end end]
}
 
puts [makeChange 100 {1 5 10 25}]
puts [makeChange 100000 {1 5 10 25 50 100}]
# Making change with the EU coin set:
puts [makeChange 100 {1 2 5 10 20 50 100 200}]
puts [makeChange 100000 {1 2 5 10 20 50 100 200}]

Output:

242
13398445413854501
4563
10056050940818192726001

uBasic/4tH[edit]

Translation of: Run BASIC
c = 0
for p = 0 to 100
for n = 0 to 20
for d = 0 to 10
for q = 0 to 4
if p + n * 5 + d * 10 + q * 25 = 100 then
print p;" pennies ";n;" nickels "; d;" dimes ";q;" quarters"
c = c + 1
endif
next q
next d
next n
next p
print c;" ways to make a buck"
Output:
0 pennies 0 nickels 0 dimes 4 quarters
0 pennies 0 nickels 5 dimes 2 quarters
0 pennies 0 nickels 10 dimes 0 quarters
...
90 pennies 2 nickels 0 dimes 0 quarters
95 pennies 1 nickels 0 dimes 0 quarters
100 pennies 0 nickels 0 dimes 0 quarters
242 ways to make a buck

0 OK, 0:312

UNIX Shell[edit]

Translation of: Common Lisp
Works with: bash
function count_change {
local -i amount=$1 coin j
local ways=(1)
shift
for coin; do
for (( j=coin; j <= amount; j++ )); do
let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
done
done
echo "${ways[amount]}"
}
count_change 100 25 10 5 1
count_change 100000 100 50 25 10 5 1
Works with: ksh version 93
function count_change {
typeset -i amount=$1 coin j
typeset ways
set -A ways 1
shift
for coin; do
for (( j=coin; j <= amount; j++ )); do
let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
done
done
echo "${ways[amount]}"
}
count_change 100 25 10 5 1
count_change 100000 100 50 25 10 5 1
Works with: ksh version 88
function count_change {
typeset -i amount=$1 coin j
typeset ways
set -A ways 1
shift
for coin; do
let j=coin
while (( j <= amount )); do
let ways[j]=${ways[j]:-0}+${ways[j-coin]:-0}
let j+=1
done
done
echo "${ways[amount]}"
}
count_change 100 25 10 5 1
# (optional task exceeds a subscript limit in ksh88)

And just for fun, here's one that works even with the original V7 shell:

Works with: sh version v7
if [ $# -lt 2 ]; then
set ${1-100} 25 10 5 1
fi
amount=$1
shift
ways_0=1
for coin in "$@"; do
j=$coin
while [ $j -le $amount ]; do
d=`expr $j - $coin`
eval "ways_$j=\`expr \${ways_$j-0} + \${ways_$d-0}\`"
j=`expr $j + 1`
done
done
eval "echo \$ways_$amount"
Output:
242
13398445413854501

VBScript[edit]

Translation of: C#
 
Function count(coins,m,n)
ReDim table(n+1)
table(0) = 1
i = 0
Do While i < m
j = coins(i)
Do While j <= n
table(j) = table(j) + table(j - coins(i))
j = j + 1
Loop
i = i + 1
Loop
count = table(n)
End Function
 
'testing
arr = Array(1,5,10,25)
m = UBound(arr) + 1
n = 100
WScript.StdOut.WriteLine count(arr,m,n)
 
Output:
242

zkl[edit]

Translation of: Scheme
fcn ways_to_make_change(x, coins=T(25,10,5,1)){
if(not coins) return(0);
if(x<0) return(0);
if(x==0) return(1);
ways_to_make_change(x, coins[1,*]) + ways_to_make_change(x - coins[0], coins)
}
ways_to_make_change(100).println();
Output:
242

Blows the stack on the optional part, so try this:

Translation of: Ruby
fcn make_change2(amount, coins){
n, m  := amount, coins.len();
table := (0).pump(n+1,List, (0).pump(m,List().write,1).copy);
foreach i,j in ([1..n],[0..m-1]){
table[i][j] = (if(i<coins[j]) 0 else table[i-coins[j]][j]) +
(if(j<1) 0 else table[i][j-1])
}
table[-1][-1]
}
 
println(make_change2( 100, T(1,5,10,25)));
make_change2(0d1000_00, T(1,5,10,25,50,100)) : "%,d".fmt(_).println();
Output:
242
13,398,445,413,854,501

ZX Spectrum Basic[edit]

Translation of: AWK

Test with emulator at full speed for reasonable performance.

10 LET amount=100
20 GO SUB 1000
30 STOP
1000 LET nPennies=amount
1010 LET nNickles=INT (amount/5)
1020 LET nDimes=INT (amount/10)
1030 LET nQuarters=INT (amount/25)
1040 LET count=0
1050 FOR p=0 TO nPennies
1060 FOR n=0 TO nNickles
1070 FOR d=0 TO nDimes
1080 FOR q=0 TO nQuarters
1090 LET s=p+n*5+d*10+q*25
1100 IF s=100 THEN LET count=count+1
1110 NEXT q
1120 NEXT d
1130 NEXT n
1140 NEXT p
1150 PRINT count
1160 RETURN