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 is my solution i do not know how efficient is it but it works,i am open for suggestion.

=0;$j--){
            $coin_count[$j]=0;
            $player[$j]=$player_copy[$j];
        }
        for($j=$z;$j>=0;$j--){
            while(($price>0) && 1<=$player[$j]){
                $price-=$coin_value[$j];
                $player[$j]--;
                $coin_count[$j]++;
            }
        }
        $change[$k++]=$price;
         if($price!=0){
                for($j=$z;$j>=0;$j--)
                    if($price_copy>$coin_value[$j]){
                        $z=$j-1;
                        $price=$price_copy;
                        goto label1;
                    }
                $flag=1;
         }
    //find minimum change 
        $minv=$change[0];

         for($i=1;$change[$i]>=0 and $i<4;$i++)
             if($change[$i]>$minv)
                $minv=$change[$i];
         $i;
  //when you find minimum change find which coins you have used
         for($i=0;$i<4;$i++)
             if($change[$i]==$minv && $flag==1){
                $flag=2;
                for($j=3;$j>=0;$j--){//reset coin_count and player budget
                    $coin_count[$j]=0;
                    $player[$j]=$player_copy[$j];
                }
                 for($j=3-($i%2)-1;$j>=0;$j--){
                     while(($price>0) && 1<=$player[$j]){
                         $price-=$coin_value[$j];
                         $player[$j]--;
                         $coin_count[$j]++;
                     }
                  }
              }
//prints result
         for($j=0;$j<4;$j++)
            printf("%d x %d\n",$coin_count[$j],$coin_value[$j]);
         printf("change: %d\n",$minv);
?>