Find minimum number of coins that make a given value
- 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
Factor
<lang factor>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 .</lang>
- Output:
{ { 200 4 } { 100 1 } { 50 1 } { 20 1 } { 10 1 } { 5 1 } { 2 1 } { 1 1 } }
Go
<lang go>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.")
}</lang>
- 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
<lang 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</lang>
- Output:
(4,200) (1,100) (1,50) (1,20) (1,10) (1,5) (1,2) (1,1)
Julia
Long version
Using a linear optimizer for this is serious overkill, but why not? <lang julia>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
</lang>
- 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))
Nim
<lang 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</lang>
- 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
Phix
with javascript_semantics sequence coins = {1,2,5,10,20,50,100,200} atom total = 988 string sc = join(apply(coins,sprint),", ") integer k = rfind(',',sc), count = 0 sc[k..k] = " and" printf(1,"Make a value of %d using the coins %s:\n",{total,sc}) 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
<lang python>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()
</lang>
- 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
<lang python>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()</lang>
- 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
Since unit denominations are possible, don't bother to check to see if an exact pay-out isn't possible.
<lang perl6>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;
}</lang>
- 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
A check was made to see if an exact pay─out isn't possible.
The total number of coins paid out is also shown. <lang rexx>/*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. */</lang>
- 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
<lang 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 </lang>
- Output:
working... Coins are: 4*200 1*100 1*50 1*20 1*10 1*5 1*2 1*1 done...
Wren
As there is, apparently, an unlimited supply of coins of each denomination, it follows that any amount can be made up. <lang ecmascript>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.")</lang>
- 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.