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

The solution I was able to made covers the 3 examples posted in your question. And always gives the change with as few coins as possible.

The tests I made seemed to be executed very fast.

Here I post the code:

= 0; $i--) {
        if (check_availability($i)) {
            return $i;
        }
    }
    return -1;
}

for($i=0;$i<4;$i++) {
    while(check_availability($i) && ($btotal + $coin_value[$i]) <= $price) {
        $btotal += $coin_value[$i];
        $barray[$i]++;
    }
}

if($price != $btotal) {
    $buf = check_lower_available();
    for ($i = $buf; $i >= 0; $i--) {
        if (check_availability($i) && ($btotal + $coin_value[$i]) > $price) {
            $btotal += $coin_value[$i];
            $barray[$i]++;
            break;
        }
    }
}

// Time to pay
$bchange = 0;
$barray_change = array(0,0,0,0);

if ($price > $btotal) {
    echo "You have not enough money.";
}
else {
    $pay_msg = "You paid $".$btotal."\n\n";
    $pay_msg.= "You used ".$barray[0]." coins of $10\n";
    $pay_msg.= "You used ".$barray[1]." coins of $5\n";
    $pay_msg.= "You used ".$barray[2]." coins of $3\n";
    $pay_msg.= "You used ".$barray[3]." coins of $1\n\n\n";
    // Time to give change
    $the_diff = $btotal - $price;
    if (!empty($the_diff)) {
        for ($i = 0; $i < 4; $i++) {
            while($the_diff >= $coin_value[$i]) {
                $bchange += $coin_value[$i];
                $barray_change[$i]++;
                $the_diff -= $coin_value[$i];
            }
        }

        $check_sum = array_sum($inventory) - array_sum($barray);
        $check_sum+= array_sum($barray_change);
        $msg = "";
        if ($check_sum < 15) {
            $change_msg = "Your change: $".$bchange."\n\n";
            $change_msg.= "You received ".$barray_change[0]." coins of $10\n";
            $change_msg.= "You received ".$barray_change[1]." coins of $5\n";
            $change_msg.= "You received ".$barray_change[2]." coins of $3\n";
            $change_msg.= "You received ".$barray_change[3]." coins of $1\n\n";
            $msg = $pay_msg.$change_msg;
        }
        else {
            $msg = "You have not enough space to hold the change.\n";
            $msg.= "Buy cancelled.\n";
        }
    }
    else {
        $msg = $pay_msg."You do not need change\n";
    }
    if ($check_sum < 15) {
        for ($i = 0; $i < 4; $i++) {
            $inventory[$i] -= $barray[$i];
            $total_coins-= $barray[$i];
        }
        for ($i = 0; $i < 4; $i++) {
            $inventory[$i] += $barray_change[$i];
            $total_coins+= $barray[$i];
        }
    }
    echo $msg;
    echo "Now you have:\n";
    echo $inventory[0]." coins of $10\n";
    echo $inventory[1]." coins of $5\n";
    echo $inventory[2]." coins of $3\n";
    echo $inventory[3]." coins of $1\n";
}