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Find minimum number of coins that make a given value

From Rosetta Code
Find minimum number of coins that make a given value is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Find and show here on this page the minimum number of coins that can make a value of   988.

Available coins are:   1,   2,   5,   10,   20,   50,   100,   and   200.


The coins that would be dispensed are:

     four coins of 200
      one coin  of 100
      one coin  of  50 
      one coin  of  20
      one coin  of  10 
      one coin  of   5 
      one coin  of   2
      one coin  of   1



APL[edit]

Works with: Dyalog APL
coins←{
{⍺,≢⍵}⌸⍺[⍒⍺]{
coin←⊃(⍵≥⍺)/⍺
coin=0:⍬
coin,⍺∇⍵-coin
}⍵
}
Output:
      (1 2 5 10 20 50 100 200) coins 988
200 4
100 1
 50 1
 20 1
 10 1
  5 1
  2 1
  1 1

AppleScript[edit]

----------------- MINIMUM NUMBER OF COINS ----------------
 
-- change :: [Int] -> Int -> [(Int, Int)]
on change(units, n)
if {} = units or 0 = n then
{}
else
set {x, xs} to {item 1 of units, rest of units}
set q to n div x
if 0 = q then
change(xs, n)
else
{{q, x}} & change(xs, n mod x)
end if
end if
end change
 
 
--------------------------- TEST -------------------------
on run
set coinReport to ¬
showChange({200, 100, 50, 20, 10, 5, 2, 1})
 
unlines(map(coinReport, {1024, 988}))
end run
 
 
-- showChange :: [Int] -> Int -> String
on showChange(units)
script
on |λ|(n)
script go
on |λ|(qd)
set {q, d} to qd
(q as text) & " * " & d as text
end |λ|
end script
unlines({("Summing to " & n as text) & ":"} & ¬
map(go, change(units, n))) & linefeed
end |λ|
end script
end showChange
 
 
------------------------- GENERIC ------------------------
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
-- The list obtained by applying f
-- to each element of xs.
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
 
 
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
 
 
-- unlines :: [String] -> String
on unlines(xs)
-- A single string formed by the intercalation
-- of a list of strings with the newline character.
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set s to xs as text
set my text item delimiters to dlm
s
end unlines
Output:
Summing to 1024:
5 * 200
1 * 20
2 * 2

Summing to 988:
4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1

F#[edit]

 
//Find minimum number of coins that make a given value - Nigel Galloway: August 12th., 20
let fN g=let rec fG n g=function h::t->fG((g/h,h)::n)(g%h) t |_->n in fG [] g [200;100;50;20;10;5;2;1]
fN 988|>List.iter(fun(n,g)->printfn "Take %d of %d" n g)
 
Output:
Take 1 of 1
Take 1 of 2
Take 1 of 5
Take 1 of 10
Take 1 of 20
Take 1 of 50
Take 1 of 100
Take 4 of 200

Factor[edit]

Works with: Factor version 0.99 2021-06-02
USING: assocs kernel math math.order prettyprint sorting ;
 
: make-change ( value coins -- assoc )
[ >=< ] sort [ /mod swap ] zip-with nip ;
 
988 { 1 2 5 10 20 50 100 200 } make-change .
Output:
{
    { 200 4 }
    { 100 1 }
    { 50 1 }
    { 20 1 }
    { 10 1 }
    { 5 1 }
    { 2 1 }
    { 1 1 }
}

Go[edit]

Translation of: Wren
package main
 
import "fmt"
 
func main() {
denoms := []int{200, 100, 50, 20, 10, 5, 2, 1}
coins := 0
amount := 988
remaining := 988
fmt.Println("The minimum number of coins needed to make a value of", amount, "is as follows:")
for _, denom := range denoms {
n := remaining / denom
if n > 0 {
coins += n
fmt.Printf("  %3d x %d\n", denom, n)
remaining %= denom
if remaining == 0 {
break
}
}
}
fmt.Println("\nA total of", coins, "coins in all.")
}
Output:
The minimum number of coins needed to make a value of 988 is as follows:
  200 x 4
  100 x 1
   50 x 1
   20 x 1
   10 x 1
    5 x 1
    2 x 1
    1 x 1

A total of 11 coins in all.

Haskell[edit]

import Data.List (mapAccumL)
import Data.Tuple (swap)
 
----------------------- FIND CHANGE ----------------------
 
change :: [Int] -> Int -> [(Int, Int)]
change xs n = zip (snd $ mapAccumL go n xs) xs
where
go m v = swap (quotRem m v)
 
 
--------------------------- TEST -------------------------
main :: IO ()
main =
mapM_ print $
change [200, 100, 50, 20, 10, 5, 2, 1] 988
Output:
(4,200)
(1,100)
(1,50)
(1,20)
(1,10)
(1,5)
(1,2)
(1,1)

Or as a hand-written recursion, defining a slightly more parsimonious listing, and allowing for denomination lists which are ill-sorted or incomplete.

import Data.List (sortOn)
import Data.Ord (Down (Down))
 
---------- MINIMUM NUMBER OF COINS TO MAKE A SUM ---------
 
change :: [Int] -> Int -> Either String [(Int, Int)]
change units n
| 0 == mod n m = Right $ go (sortOn Down units) (abs n)
| otherwise =
Left $
concat
[ "Residue of ",
show (mod n m),
" - no denomination smaller than ",
show m,
"."
]
where
m = minimum units
go _ 0 = []
go [] _ = []
go (x : xs) n
| 0 == q = go xs n
| otherwise = (q, x) : go xs r
where
(q, r) = quotRem n x
 
--------------------------- TEST -------------------------
main :: IO ()
main = mapM_ putStrLn $ test <$> [1024, 988]
where
test n =
either
id
( concat
. (:) ("Summing to " <> show n <> ":\n")
. fmap
( \(q, v) ->
concat
[show q, " * ", show v, "\n"]
)
)
(change [200, 100, 50, 20, 10, 5, 2, 1] n)
Output:
Summing to 1024:
5 * 200
1 * 20
2 * 2

Summing to 988:
4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1

JavaScript[edit]

Works with: JavaScript version ES6
(() => {
"use strict";
 
// -- MINIMUM NUMBER OF COINS TO MAKE A GIVEN VALUE --
 
// change :: [Int] -> Int -> [(Int, Int)]
const change = denominations => {
// A minimum list of (quantity, value) pairs for n.
// Unused denominations are excluded from the list.
const go = n => {
const m = abs(n);
 
return 0 < denominations.length && 0 < m ? (
() => {
const [h, ...t] = denominations;
const q = Math.trunc(m / h);
 
return (
0 < q ? [
[q, h]
] : []
).concat(change(t)(m % h));
}
)() : [];
};
 
return go;
};
 
 
// ---------------------- TEST -----------------------
// main :: IO ()
const main = () => {
// Two sums tested with a set of denominations.
const f = change([200, 100, 50, 20, 10, 5, 2, 1]);
 
return [1024, 988].reduce((acc, n) => {
const
report = f(n).reduce(
(a, [q, u]) => `${a}${q} * ${u}\n`,
""
);
 
return `${acc}Summing to ${abs(n)}:\n` + (
`${report}\n`
);
},
""
);
};
 
 
// --------------------- GENERIC ---------------------
 
// abs :: Num -> Num
const abs =
// Absolute value - without the sign.
x => 0 > x ? (
-x
) : x;
 
 
// MAIN ---
return main();
})();
Output:
Summing to 1024:
5 * 200
1 * 20
2 * 2

Summing to 988:
4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1

jq[edit]

Works with: jq

Works with gojq, the Go implementation of jq

 
# If $details then provide {details, coins}, otherwise just the number of coins.
def minimum_number($details):
. as $amount
| [200, 100, 50, 20, 10, 5, 2, 1] as $denoms
| {coins: 0, remaining: 988, details: []}
| label $out
| foreach $denoms[] as $denom (.;
((.remaining / $denom)|floor) as $n
| if $n > 0
then .coins += $n
| if $details then .details += [{$denom, $n}] else . end
| .remaining %= $denom
else . end;
if .remaining == 0 then ., break $out else empty end)
| if $details then {details, coins} else .coins end ;
 
# Verbose mode:
def task:
"\nThe minimum number of coins needed to make a value of \(.) is as follows:",
(minimum_number(true)
| .details[],
"\nA total of \(.coins) coins in all." );
 
 
988
| minimum_number(false), # illustrate minimal output
task # illustrate detailed output
 
Output:
11

The minimum number of coins needed to make a value of 988 is as follows:
{"denom":200,"n":4}
{"denom":100,"n":1}
{"denom":50,"n":1}
{"denom":20,"n":1}
{"denom":10,"n":1}
{"denom":5,"n":1}
{"denom":2,"n":1}
{"denom":1,"n":1}

A total of 11 coins in all.

</pre>

Julia[edit]

Long version[edit]

Using a linear optimizer for this is serious overkill, but why not?

using JuMP, GLPK
 
model = Model(GLPK.Optimizer)
@variable(model, ones, Int)
@variable(model, twos, Int)
@variable(model, fives, Int)
@variable(model, tens, Int)
@variable(model, twenties, Int)
@variable(model, fifties, Int)
@variable(model, onehundreds, Int)
@variable(model, twohundreds, Int)
@constraint(model, ones >= 0)
@constraint(model, twos >= 0)
@constraint(model, fives >= 0)
@constraint(model, tens >= 0)
@constraint(model, twenties >= 0)
@constraint(model, fifties >= 0)
@constraint(model, onehundreds >= 0)
@constraint(model, twohundreds >= 0)
@constraint(model, 988 == 1ones +2twos + 5fives + 10tens + 20twenties + 50fifties + 100onehundreds + 200twohundreds)
 
@objective(model, Min, ones + twos + fives + tens + twenties + fifties + onehundreds + twohundreds)
 
optimize!(model)
println("Optimized total coins: ", objective_value(model))
for val in [ones, twos, fives, tens, twenties, fifties, onehundreds, twohundreds]
println("Value of ", string(val), " is ", value(val))
end
 
Output:
Optimized total coins: 11.0
Value of ones is 1.0
Value of twos is 1.0
Value of fives is 1.0
Value of tens is 1.0
Value of twenties is 1.0
Value of fifties is 1.0
Value of onehundreds is 1.0
Value of twohundreds is 4.0

Brief REPL command version[edit]

julia> accumulate((x, y) -> (x[1] % y, (y, x[1] ÷ y)), [200, 100, 50, 20, 10, 5, 2, 1], init=(988, 0))
8-element Vector{Tuple{Int64, Tuple{Int64, Int64}}}:
 (188, (200, 4))
 (88, (100, 1))
 (38, (50, 1))
 (18, (20, 1))
 (8, (10, 1))
 (3, (5, 1))
 (1, (2, 1))
 (0, (1, 1))

Mathematica/Wolfram Language[edit]

coins = {1, 2, 5, 10, 20, 50, 100, 200};
out = v /. ConvexOptimization[Total[v], coins . v == 988, v \[Element] Vectors[8, NonNegativeIntegers]];
MapThread[Row[{#1, " x ", #2}] &, {out, coins}] // Column
Output:
1 x 1
1 x 2
1 x 5
1 x 10
1 x 20
1 x 50
1 x 100
4 x 200

MiniZinc[edit]

 
%Find minimum number of coins that make a given value. Nigel Galloway, August 11th., 2021
int: N=988;
array [1..8] of int: coinValue=[1,2,5,10,20,50,100,200];
array [1..8] of var 0..N: take; constraint sum(n in 1..8)(take[n]*coinValue[n])=N;
solve minimize sum(n in 1..8)(take[n]);
output(["Take "++show(take[n])++" of "++show(coinValue[n])++"\n" | n in 1..8])
 
Output:
Take 1 of 1
Take 1 of 2
Take 1 of 5
Take 1 of 10
Take 1 of 20
Take 1 of 50
Take 1 of 100
Take 4 of 200
----------
==========
Finished in 196msec

Nim[edit]

import strformat
 
const
Coins = [200, 100, 50, 20, 10, 5, 2, 1]
Target = 988
 
echo &"Minimal number of coins to make a value of {Target}:"
var count = 0
var remaining = Target
for coin in Coins:
let n = remaining div coin
if n != 0:
inc count, n
echo &"coins of {coin:3}: {n}"
dec remaining, n * coin
if remaining == 0: break
 
echo "\nTotal: ", count
Output:
Minimal number of coins to make a value of 988:
coins of 200: 4
coins of 100: 1
coins of  50: 1
coins of  20: 1
coins of  10: 1
coins of   5: 1
coins of   2: 1
coins of   1: 1

Total: 11

Perl[edit]

use strict;
use warnings;
 
my @denominations = <200 100 50 20 10 5 2 1>;
 
sub change {
my $n = shift;
my @a;
push(@a, int $n/$_) and $n %= $_ for @denominations;
@a
}
 
my @amounts = change 988;
for (0 .. $#amounts) {
printf "%1d * %3d\n", $amounts[$_], $denominations[$_]
}
Output:
4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1

Phix[edit]

with javascript_semantics
requires("1.0.1") -- (lastdelim added to the join() function)
sequence coins = {1,2,5,10,20,50,100,200}
string strc = join(apply(coins,sprint),", ", ", and ")
atom total = 988
printf(1,"Make a value of %d using the coins %s:\n",{total,strc})
integer count = 0
for i=length(coins) to 1 by -1 do
    integer ci = coins[i],
            c = floor(total/ci)
    if c then
        printf(1,"%6s coin%s of %3d\n",{ordinal(c,true),iff(c>1?"s":" "),ci})
        count += c
        total = remainder(total,ci)
        if total=0 then exit end if
    end if
end for
printf(1,"%s coins were used.\n",{proper(ordinal(count,true))})
Output:
Make a value of 988 using the coins 1, 2, 5, 10, 20, 50, 100, and 200:
  four coins of 200
   one coin  of 100
   one coin  of  50
   one coin  of  20
   one coin  of  10
   one coin  of   5
   one coin  of   2
   one coin  of   1
Eleven coins were used.

Python[edit]

Python :: Procedural[edit]

def makechange(denominations = [1,2,5,10,20,50,100,200], total = 988):
print(f"Available denominations: {denominations}. Total is to be: {total}.")
coins, remaining = sorted(denominations, reverse=True), total
for n in range(len(coins)):
coinsused, remaining = divmod(remaining, coins[n])
if coinsused > 0:
print(" ", coinsused, "*", coins[n])
 
makechange()
 
Output:
Available denominations: [1, 2, 5, 10, 20, 50, 100, 200]. Total is to be: 988.
    4 * 200
    1 * 100
    1 * 50
    1 * 20
    1 * 10
    1 * 5
    1 * 2
    1 * 1

Python :: Functional[edit]

'''Minimum number of coins to make a given value'''
 
 
# change :: [Int] -> Int -> [(Int, Int)]
def change(xs):
'''A list of (quantity, denomination) pairs.
Unused denominations are excluded from the list.
'''

def go(n):
if xs and n:
h, *t = xs
q, r = divmod(n, h)
 
return ([(q, h)] if q else []) + (
change(t)(r)
)
else:
return []
 
return go
 
 
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Testing a set of denominations with two sums'''
 
f = change([200, 100, 50, 20, 10, 5, 2, 1])
print(
"\n".join([
f'Summing to {n}:\n' + "\n".join([
f'{qu[0]} * {qu[1]}' for qu in f(n)]
) + "\n"
for n in [1024, 988]
])
)
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
Summing to 1024:
5 * 200
1 * 20
2 * 2

Summing to 988:
4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1

Raku[edit]

Since unit denominations are possible, don't bother to check to see if an exact pay-out isn't possible.

my @denominations = 200, 100, 50, 20, 10, 5, 2, 1;
 
sub change (Int $n is copy where * >= 0) { gather for @denominations { take $n div $_; $n %= $_ } }
 
for 988, 1307, 37511, 0 -> $amount {
say "\n$amount:";
printf "%d × %d\n", |$_ for $amount.&change Z @denominations;
}
Output:
988:
4 × 200
1 × 100
1 × 50
1 × 20
1 × 10
1 × 5
1 × 2
1 × 1

1307:
6 × 200
1 × 100
0 × 50
0 × 20
0 × 10
1 × 5
1 × 2
0 × 1

37511:
187 × 200
1 × 100
0 × 50
0 × 20
1 × 10
0 × 5
0 × 2
1 × 1

0:
0 × 200
0 × 100
0 × 50
0 × 20
0 × 10
0 × 5
0 × 2
0 × 1

REXX[edit]

A check was made to see if an exact pay─out isn't possible.

The total number of coins paid out is also shown.

/*REXX pgm finds & displays the minimum number of coins which total to a specified value*/
parse arg $ coins /*obtain optional arguments from the CL*/
if $='' | $="," then $= 988 /*Not specified? Then use the default.*/
if coins='' | coins="," then coins= 1 2 5 10 20 50 100 200 /* ... " " " " */
#= words(coins) /*#: is the number of allowable coins.*/
w= 0 /*width of largest coin (for alignment)*/
do j=1 for #; @.j= word(coins, j) /*assign all coins to an array (@.). */
w= max(w, length(@.j) ) /*find the width of the largest coin. */
end /*j*/
say 'available coin denominations: ' coins /*shown list of available denominations*/
say
say center(' making change for ' $, 30 ) /*display title for the upcoming output*/
say center('' , 30, "─") /* " sep " " " " */
koins= 0 /*the total number of coins dispensed. */
paid= 0 /*the total amount of money paid so far*/
do k=# by -1 for #; z= $ % @.k /*start with largest coin for payout. */
if z<1 then iterate /*if Z is not positive, then skip coin.*/
koins= koins + z
paid= z * @.k /*pay out a number of coins. */
$= $ - paid /*subtract the pay─out from the $ total*/
say right(z,9) ' of coin ' right(@.k, w) /*display how many coins were paid out.*/
end /*k*/
 
say center('' , 30, "─") /* " sep " " " " */
say
say 'number of coins dispensed: ' koins
if $>0 then say 'exact payout not possible.' /*There a residue? Payout not possible*/
exit 0 /*stick a fork in it, we're all done. */
output   when using the default inputs:
available coin denominations:  1 2 5 10 20 50 100 200

    making change for  988
──────────────────────────────
        4  of coin  200
        1  of coin  100
        1  of coin   50
        1  of coin   20
        1  of coin   10
        1  of coin    5
        1  of coin    2
        1  of coin    1
──────────────────────────────

number of coins dispensed:  11

Ring[edit]

 
load "stdlib.ring"
 
see "working..." + nl
see "Coins are:" + nl
sum = 988
 
sumCoins = 0
coins = [1,2,5,10,20,50,100,200]
coins = reverse(coins)
 
for n = 1 to len(coins)
nr = floor(sum/coins[n])
if nr > 0
sumCoins= nr*coins[n]
sum -= sumCoins
see "" + nr + "*" + coins[n] + nl
ok
next
 
see "done..." + nl
 
Output:
working...
Coins are:
4*200
1*100
1*50
1*20
1*10
1*5
1*2
1*1
done...


Rust[edit]

 
fn main() {
let denoms = vec![200, 100, 50, 20, 10, 5, 2, 1];
let mut coins = 0;
let amount = 988;
let mut remaining = 988;
println!("The minimum number of coins needed to make a value of {} is as follows:", amount);
for denom in denoms.iter() {
let n = remaining / denom;
if n > 0 {
coins += n;
println!(" {} x {}", denom, n);
remaining %= denom;
if remaining == 0 {
break;
}
}
}
println!("\nA total of {} coins in all.", coins);
}
 
Output:
The minimum number of coins needed to make a value of 988 is as follows:
  200 x 4
  100 x 1
  50 x 1
  20 x 1
  10 x 1
  5 x 1
  2 x 1
  1 x 1

A total of 11 coins in all.

Wren[edit]

Library: Wren-fmt

As there is, apparently, an unlimited supply of coins of each denomination, it follows that any amount can be made up.

import "/fmt" for Fmt
 
var denoms = [200, 100, 50, 20, 10, 5, 2, 1]
var coins = 0
var amount = 988
var remaining = 988
System.print("The minimum number of coins needed to make a value of %(amount) is as follows:")
for (denom in denoms) {
var n = (remaining / denom).floor
if (n > 0) {
coins = coins + n
Fmt.print(" $3d x $d", denom, n)
remaining = remaining % denom
if (remaining == 0) break
}
}
System.print("\nA total of %(coins) coins in all.")
Output:
The minimum number of coins needed to make a value of 988 is as follows:
  200 x 4
  100 x 1
   50 x 1
   20 x 1
   10 x 1
    5 x 1
    2 x 1
    1 x 1

A total of 11 coins in all.

XPL0[edit]

Translation of: Wren
int Denom, Denoms, Coins, Amount, Remaining, I, N;
[Denoms:= [200, 100, 50, 20, 10, 5, 2, 1];
Coins:= 0;
Amount:= 988;
Remaining:= 988;
Text(0, "The minimum number of coins needed to make a value of ");
IntOut(0, Amount); Text(0, " is as follows:
");
Format(3, 0);
for I:= 0 to 7 do
[Denom:= Denoms(I);
N:= Remaining/Denom;
if N > 0 then
[Coins:= Coins + N;
RlOut(0, float(Denom)); Text(0, " x "); IntOut(0, N); CrLf(0);
Remaining:= rem(Remaining/Denom);
if Remaining = 0 then I:= 7;
];
];
Text(0, "
A total of "); IntOut(0, Coins); Text(0, " coins in all.
");
]
Output:
The minimum number of coins needed to make a value of 988 is as follows:
200 x 4
100 x 1
 50 x 1
 20 x 1
 10 x 1
  5 x 1
  2 x 1
  1 x 1

A total of 11 coins in all.