Find minimum number of coins that make a given value

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

11l

Translation of: Python_%3A%3A_Procedural

V denominations = [1, 2, 5, 10, 20, 50, 100, 200]
V total = 988
print(‘Available denominations: ’denominations‘. Total is to be: ’total‘.’)
V (coins, remaining) = (sorted(denominations, reverse' 1B), total)
L(n) 0 .< coins.len
   (V coinsused, remaining) = divmod(remaining, coins[n])
   I coinsused > 0
      print(‘    ’coinsused‘ * ’coins[n])

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

Action!

PROC Main()
  DEFINE LEN="8"
  BYTE ARRAY coins=[200 100 50 20 10 5 2 1],count(LEN)
  BYTE i
  INT value=[988],curr,total

  Zero(count,LEN)
  i=0 total=0
  curr=value
  WHILE curr>0
  DO
    IF curr>=coins(i) THEN
      count(i)==+1
      total==+1
      curr==-coins(i)
    ELSE
      i==+1
    FI
  OD

  PrintF("%I coins to make %I:%E",total,value)
  FOR i=0 TO LEN-1
  DO
    IF count(i) THEN
      PrintF("  %B x %B%E",count(i),coins(i))
    FI
  OD
RETURN

Output:

Screenshot from Atari 8-bit computer

11 coins to make 988:
  4 x 200
  1 x 100
  1 x 50
  1 x 20
  1 x 10
  1 x 5
  1 x 2
  1 x 1

ALGOL 68

Translation of: Wren

BEGIN # find the minimum number of coins needed to make a given value         #
      # translated from the Wren sample                                       #

    []INT denoms     = ( 200, 100, 50, 20, 10, 5, 2, 1 );
    INT   coins     := 0;
    INT   amount     = 988;
    INT   remaining := amount;
    print( ( "The minimum number of coins needed to make a value of " ) );
    print( ( whole( amount, 0 ), " is as follows:", newline ) );
    FOR d pos FROM LWB denoms TO UPB denoms
    WHILE INT denom = denoms[ d pos ];
          INT n     = remaining OVER denom;
          IF n > 0 THEN
              coins +:= n;
              print( ("  ", whole( denom, -3 ), " x ", whole( n, 0 ), newline ) );
              remaining MODAB denom
          FI;
          remaining > 0
    DO SKIP OD;
    print( ( newline, "A total of ", whole( coins, 0 ), " coins in all.", newline ) )
END

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.

APL

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

----------------- 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

Arturo

coins: [200 100 50 20 10 5 2 1]
target: 988

print ["Minimum number of coins to make a value of " (to :string target)++":"]

cnt: 0
remaining: new target

loop coins 'coin [
    n: remaining / coin
    if not? zero? n [
        cnt: cnt + n
        print ["    coins of" coin "->" n]
        remaining: remaining - n * coin
        if zero? remaining -> break
    ]
]

print ["\nTotal: " cnt]

Output:

Minimum 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

AutoHotkey

coins := [1, 2, 5, 10, 20, 50, 100, 200]
val := 988

result := ""
while val
{
    coin := coins.pop()
    if (val//coin)
        result .= val//coin " * " coin "`n", val -= val//coin * coin
}
MsgBox, 262144, , % result
return

Output:

4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1

AWK

# syntax: GAWK -f FIND_MINIMUM_NUMBER_OF_COINS_THAT_MAKE_A_GIVEN_VALUE.AWK
BEGIN {
    n = split("200,100,50,20,10,5,2,1",arr,",")
    main(988)
    main(388)
    main(0)
    exit(0)
}
function main(arg1,  amount,coins,denomination,i,remaining,total) {
    amount = remaining = int(arg1)
    for (i=1; i<=n; i++) {
      denomination = arr[i]
      coins = 0
      while (remaining >= denomination) {
        remaining -= denomination
        coins++
      }
      total += coins
      printf("%4d x %2d = %d\n",denomination,coins,denomination*coins)
    }
    printf("%9d coins needed to disperse %s\n\n",total,arg1)
}

Output:

 200 x  4 = 800
 100 x  1 = 100
  50 x  1 = 50
  20 x  1 = 20
  10 x  1 = 10
   5 x  1 = 5
   2 x  1 = 2
   1 x  1 = 1
       11 coins needed to disperse 988

 200 x  1 = 200
 100 x  1 = 100
  50 x  1 = 50
  20 x  1 = 20
  10 x  1 = 10
   5 x  1 = 5
   2 x  1 = 2
   1 x  1 = 1
        8 coins needed to disperse 388

 200 x  0 = 0
 100 x  0 = 0
  50 x  0 = 0
  20 x  0 = 0
  10 x  0 = 0
   5 x  0 = 0
   2 x  0 = 0
   1 x  0 = 0
        0 coins needed to disperse 0

BASIC256

amount = 988
sumCoins = 0
dim coins = {1, 2, 5, 10, 20, 50, 100, 200}

print "Make a value of "; amount; " using the coins 1, 2, 5, 10, 20, 50, 100 and 200:"

for n = coins[?]-1 to 0 step -1
	tmp = floor(amount/coins[n])
	if tmp >= 0 then
		print tmp; " * "; coins[n]
		sumCoins = sumCoins + tmp
		amount = amount % coins[n]
	end if
next n
end

Output:

Igual que la entrada de FreeBASIC.

C

#include <stdio.h>

#define TOTAL    988
#define Q_VALUES   8

int main() {
    const int kValues[Q_VALUES] = { 200, 100, 50, 20, 10, 5, 2, 1 };
    int t, q, iv;

    for( t=TOTAL, iv=0; iv<Q_VALUES; t%=kValues[iv], ++iv ) {
        q = t/kValues[iv];
        printf( "%4d coin%c of %4d\n", q, q!=1?'s':' ', kValues[iv] );
    }

    return 0;
}

Output:

   4 coins of  200
   1 coin  of  100
   1 coin  of   50
   1 coin  of   20
   1 coin  of   10
   1 coin  of    5
   1 coin  of    2
   1 coin  of    1

Delphi

Works with: Delphi version 6.0

Library: SysUtils,StdCtrls

const Coins: array [0..7] of integer = (1,2,5,10,20,50,100,200);

procedure MinimumCoins(Memo: TMemo; Value: integer);
var I,C: integer;
begin
Memo.Lines.Add('Providing Change for: '+IntToStr(Value));
for I:=High(Coins) downto 0 do
	begin
	C:=Value div Coins[I];
	Value:=Value mod Coins[I];
	Memo.Lines.Add(IntToStr(C)+' coins of '+IntToStr(Coins[I]));
	end;
Memo.Lines.Add('');
end;


procedure TestMinimumCoins(Memo: TMemo);
begin
MinimumCoins(Memo,988);
MinimumCoins(Memo,1307);
MinimumCoins(Memo,37511);
MinimumCoins(Memo,0);
end;

Output:

Providing Change for: 988
4 coins of 200
1 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

Providing Change for: 1307
6 coins of 200
1 coins of 100
0 coins of 50
0 coins of 20
0 coins of 10
1 coins of 5
1 coins of 2
0 coins of 1

Providing Change for: 37511
187 coins of 200
1 coins of 100
0 coins of 50
0 coins of 20
1 coins of 10
0 coins of 5
0 coins of 2
1 coins of 1

Providing Change for: 0
0 coins of 200
0 coins of 100
0 coins of 50
0 coins of 20
0 coins of 10
0 coins of 5
0 coins of 2
0 coins of 1

EasyLang

sum = 988
coins[] = [ 200 100 50 20 10 5 2 1 ]
# 
for coin in coins[]
   n = sum div coin
   if n > 0
      print "coins of " & coin & ": " & n
      sum -= n * coin
   .
.

Output:

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

F#

//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

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 }
}

FreeBASIC

#define floor(x) ((x*2.0-0.5) Shr 1)

Dim As Integer amount = 988
Dim As Integer sumCoins = 0
Dim As Integer n, tmp
Dim As Integer coins(8) = {1, 2, 5, 10, 20, 50, 100, 200}

Print "Make a value of"; amount; " using the coins 1, 2, 5, 10, 20, 50, 100 and 200:"

For n As Integer = Ubound(coins) To 0 Step -1
    tmp = floor(amount/coins(n))
    If tmp >= 0 Then
        Print tmp; " *"; coins(n)
        sumCoins += tmp
        amount Mod= coins(n)
    End If
Next n
Sleep

Output:

Make a value of 988 using the coins 1, 2, 5, 10, 20, 50, 100 and 200:
 4 * 200
 1 * 100
 1 * 50
 1 * 20
 1 * 10
 1 * 5
 1 * 2
 1 * 1

FutureBasic

Task solution wrapped into a general purpose function with test examples shown.

void local fn MinimumCoinsForValue( value as NSUInteger, coins as CFArrayRef )
  NSUInteger  i, count, tmp
  CFStringRef coinStr = fn ArrayComponentsJoinedByString( coins, @", " )
  
  printf @"The minimum number of coins valued %@ needed to total %lu is:", coinStr, value
  
  count = len(coins)
  for i = count to 1 step -1
    tmp = (NSUInteger)fn floor( value / fn NumberIntegerValue( coins[i-1] ) )
    if ( tmp > 0 )
      printf @"%lu * %@", tmp, coins[i-1]
      value = value mod fn NumberIntegerValue( coins[i-1] )
    end if
  next
end fn

fn MinimumCoinsForValue( 988, @[@1, @2, @5, @10, @20, @50, @100, @200] )
print : print
fn MinimumCoinsForValue( 1024, @[@1, @2, @5, @10, @20, @50, @100, @200] )
print : print
print "Currency in this example represents U.S. denominations:"
print "   1 cent, 5 cents, 10 cents, 25 cents, 50 cents, $1, $5, $10, $20, $50"
fn MinimumCoinsForValue( 65273, @[@1, @5, @10, @25, @50, @100, @500, @1000, @2000, @5000] )

HandleEvents

Output:

The minimum number of coins valued 1, 2, 5, 10, 20, 50, 100, 200 needed to total 988 is:
4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1


The minimum number of coins valued 1, 2, 5, 10, 20, 50, 100, 200 needed to total 1024 is:
5 * 200
1 * 20
2 * 2


Currency in this example represents U.S. denominations:
   1 cent, 5 cents, 10 cents, 25 cents, 50 cents, $1, $5, $10, $20, $50
The minimum number of coins valued 1, 5, 10, 25, 50, 100, 500, 1000, 2000, 5000 needed to total 65273 is:
13 * 5000
2 * 100
1 * 50
2 * 10
3 * 1

Go

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

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

J

coins=. [ (,: {."1@}:) [: (#: {:)/\. 0 ,. ,

1 2 5 10 20 50 100 200 coins 988

Output:

1 2 5 10 20 50 100 200
1 1 1  1  1  1   1   4

JavaScript

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

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.

Julia

Long version

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

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))

Lua

Translation of: Wren

do -- find the minimum number of coins needed to make a given value
   -- translated from the Wren sample

    local denoms = { 200, 100, 50, 20, 10, 5, 2, 1 }
    local amount = 988;
    local coins, remaining = 0, amount
    print( "The minimum number of coins needed to make a value of "..amount.." is as follows:" )
    for _, denom in pairs( denoms ) do
        local n = math.floor( remaining / denom )
        if n > 0 then
            coins = coins + n
            print( string.format( "%6d", denom ).." x "..n )
            remaining = remaining % denom
        end
        if remaining == 0 then break end
    end
    print( "A total of "..coins.." coins in all." )
end

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.

Mathematica/Wolfram Language

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

%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

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

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

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

Python :: Procedural

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

'''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

Quackery

  [ ' [ 200 100 50 20 10 5 2 1 ] ] is coins (   --> [ )

  [ [] swap
    coins witheach
      [ /mod dip join ]
    drop
    witheach
      [ dup 0 > iff
          [ echo say " * "
            coins i^ peek echo
            cr ]
        else drop ] ]              is task  ( n -->   )

  ' [ 988 345 1024 ]
  witheach
    [ say "To make "
      dup echo say ":" cr
      task cr ]

Output:

To make 988:
4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1

To make 345:
1 * 200
1 * 100
2 * 20
1 * 5

To make 1024:
5 * 200
1 * 20
2 * 2

Raku

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

Red

Red[]

value: 988
foreach denomination [200 100 50 20 10 5 2 1][
    quantity: to-integer value / denomination
    unless 0 = quantity [print [quantity "*" denomination]]
    value: value % denomination
]

Output:

4 * 200
1 * 100
1 * 50
1 * 20
1 * 10
1 * 5
1 * 2
1 * 1

REXX

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

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...

RPL

Works with: HP version 48G

« { 200 100 50 20 10 5 2 1 } { } 
  → coinset result
   « 1 coinset SIZE FOR j
        coinset j GET
        MOD LASTARG / IP
        'result' SWAP STO+
     NEXT
     DROP result DUP ∑LIST "coins" →TAG
» » 'COINS' STO

988 COINS

Output:

2: { 4 1 1 1 1 1 1 1 }
1: coins: 11

Rust

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.

V (Vlang)

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

Wren

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

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.

Yabasic

Translation of: FreeBASIC

amount = 988
sumCoins = 0
dim coins(7)
coins(0) = 1
coins(1) = 2
coins(2) = 5
coins(3) = 10
coins(4) = 20
coins(5) = 50
coins(6) = 100
coins(7) = 200

print "Make a value of ", amount, " using the coins 1, 2, 5, 10, 20, 50, 100 and 200:"

for n = arraysize(coins(),1) to 0 step -1
    tmp = floor(amount/coins(n))
    if tmp >= 0 then
        print tmp, " * ", coins(n)
        sumCoins = sumCoins + tmp
        amount = mod(amount, coins(n))
    end if
next n
end

Output:

Igual que la entrada de FreeBASIC.