Comments on: Code Challenge: Change Dispenser

By: John Vivenzio

John Vivenzio — Sat, 02 May 2009 06:27:15 +0000

Obscure bug?

When testing just now I found if you enter makeChange(279.96 * 100) – it rounds to 27995.9999 when multiplied by 100 – very bizarre. …using Firefox 3.5b4 – reminds me of the Pentium 59.999… days.

By: John Vivenzio

John Vivenzio — Sat, 02 May 2009 05:58:45 +0000

Last One... with inventory of bills and coins and shows remainder if can't make exact change. var b = [ // checkInventory to null out what's not available (on hand is b[i][1]) [10000, 0, "Hundred"], [ 5000, 1, "Fifty", "Fifties"], [ 2000, 0, "Twenty", "Twenties"], [ 1000, 0, "Ten"], [ 500, 0, "Five"], [ 200, 0, "Two"], [ 100, 0, "Dollar"], [ 25, 0, "quarter"], [ 10, 0, "dime"], [ 5, 0, "nickel"], [ 1, 0, "penny", "pennies"], [ 0] ]; function makeChange(money) { if (b[i][1] && (x = Math.floor(money / b[i][0]))) output.push(x + " " + (x <= 1 ? b[i][2] : !b[i][3] ? b[i][2] + "s" : b[i][3])); return isNaN(b[i + 1]) ? makeChange(b[i][1] ? money % b[i++][0] : (i++, money)) : output.push("Remainder: $" + money / 100), output; } makeChange(123279.96 * 100).join("\n");

By: John Vivenzio

John Vivenzio — Sat, 02 May 2009 04:04:34 +0000

Looking at the length of the other solutions around on here and the web (didn’t see them before and never realized this is a famous problem) I think that code I just posted is

LOL

Great blog by the way!

By: John Vivenzio

John Vivenzio — Sat, 02 May 2009 03:52:03 +0000

Didn’t think about that but just saw the ATM problem – that can be solved by running a checkInventory(array) on the array which nulls out or flags any change not available (in the above recursive) so the algorithm is easily adaptable.

-John

By: John Vivenzio

John Vivenzio — Sat, 02 May 2009 03:43:25 +0000

Here’s a quick 2 line recursive I threw together… in the name of silly frugality – I just put the names that won’t take an ’s’ in the [2] part of the array. : )

var n = [
[10000, "Hundred"],
[ 5000, "Fifty", "Fifties"],
[ 2000, "Twenty", "Twenties"],
[ 1000, "Ten"],
[ 500, "Five"],
[ 200, "Two"],
[ 100, "Dollar"],
[ 25, "quarter"],
[ 10, "dime"],
[ 5, "nickel"],
[ 1, "penny", "pennies"],
[ 0]
];

var output = [];
var i = 0, x = 0;

function makeChange(money) {
if (x = Math.floor(money / n[i][0]))
output.push(x + ” ” + (x <= 1 ? n[i][1] : !n[i][2] ? n[i][1] + “s” : n[i][2]));
return isNaN(n[i + 1]) ? makeChange(money % n[i++][0]) : output;
}

makeChange(123279.96 * 100).join(“\n”);