Choosing coins with least or no change given

Question

I am making a game which consists of coin denominations of $10, $5, $3, and $1. The player may have 0 or more of each type of currency in his inventory with a maximum of 15 coin

Answer 1

This answer is based off of גלעד-ברקן's answer. I am posting it here as per his request. While none of the answers were quite the one that I was looking for I found that this was the best option posted. Here is the modified algorithm that I am currently using:

 $have){
            $error = ["error", "Insufficient Funds"];
            return $error;
    }

    $stack = [[0,$price,$have,[]]];
    $best = [-max($coin_value),[]];

    while (!empty($stack)){

            // each stack call traverses a different set of parameters
            $parameters = array_pop($stack);

            $i = $parameters[0];
            $owed = $parameters[1];
            $have = $parameters[2];
            $result = $parameters[3];

            if ($owed <= 0){
                    //NOTE: check for new option with least change OR if same as previous option check which uses the least coins paid
                    if ($owed > $best[0] || ($owed == $best[0] && (array_sum($result) < array_sum($best[1])))){

                            //NOTE: add extra zeros to end if needed
                            while (count($result) < 4){
                                    $result[] = 0;
                            }
                            $best = [$owed,$result];
                    }
                    continue;
            }

            // skip if we have none of this coin
            if ($inventory[$i] == 0){
                    $result[] = 0;
                    $stack[] = [$i + 1,$owed,$have,$result];
                    continue;
            }

            // minimum needed of this coin
            $need = $owed - $have + $inventory[$i] * $coin_value[$i];

            if ($need < 0){
                    $min = 0;
            } else {
                    $min = ceil($need / $coin_value[$i]);
            }

            // add to stack
            for ($j=$min; $j<=$inventory[$i]; $j++){
                    $stack[] = [$i + 1,$owed - $j * $coin_value[$i],$have - $inventory[$i] * $coin_value[$i],array_merge($result,[$j])];
                    if ($owed - $j * $coin_value[$i] < 0){
                            break;
                    }
            }
    }

    return $best;
}

Here is my test code:

$start = microtime(true);

$inventory = [0,1,3,4];
$price = 12;
echo "\n";
echo json_encode(leastChange($inventory,$price));
echo "\n";



$inventory = [0,1,4,0];
$price = 12;
echo "\n";
echo json_encode(leastChange($inventory,$price));
echo "\n";

$inventory = [0,1,4,0];
$price = 6;
echo "\n";
echo json_encode(leastChange($inventory,$price));
echo "\n";


$inventory = [0,1,4,0];
$price = 7;
echo "\n";
echo json_encode(leastChange($inventory,$price));
echo "\n";


$inventory = [1,3,3,10];
$price=39;
echo "\n";
echo json_encode(leastChange($inventory,$price));
echo "\n";

$inventory = [1,3,3,10];
$price=45;
echo "\n";
echo json_encode(leastChange($inventory,$price));
echo "\n";

//stress test
$inventory = [25,25,25,1];
$price=449;
echo "\n";
echo json_encode(leastChange($inventory,$price));
echo "\n";

$time_elapsed = microtime(true) - $start;
echo "\n Time taken: $time_elapsed \n";

The result:

[0,[0,1,2,1]]

[0,[0,0,4,0]]

[0,[0,0,2,0]]

[-1,[0,1,1,0]]

[0,[1,3,3,5]]

["error","Insufficient Funds"]

[-1,[25,25,25,0]]

Time taken: 0.0046839714050293

Of course that time is in microseconds and therefore it executed in a fraction of a second!