In the game Pickomino, there are several tiles lying in the middle of the table, each with a different positive integer on them. Each turn, the players roll dices in a certain way and get a score, which is a nonnegative integer.

Now the player takes the tile with the highest number that is still lower or equal to their score, removing the tile from the middle and adding it to their stack. If this is not possible because all numbers in the middle are higher than the player's score, the player loses the topmost tile from their stack (which was added latest), which is returned to the middle. If the player has no tiles left, nothing happens.

The challenge

Simulate a player playing the game against themselves. You get a list of the tiles in the middle and a list of the scores that the player got. Return a list of the tiles of the player after all turns have been evaluated.

Challenge rules

• You can assume that the list with the tiles is ordered and doesn't contain any integer twice.


• You can take both lists of input in any order you want


• The output has to keep the order of the tiles on the stack, but you can decide whether the list is sorted from top to bottom or from bottom to top.

General rules

• This is code-golf, so shortest answer in bytes wins.
  
  Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.


• 
  Standard rules apply for your answer with default I/O rules, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs.


• 
  Default Loopholes are forbidden.


• If possible, please add a link with a test for your code (i.e. TIO).


• Adding an explanation for you answer is recommended.

Example

(taken from the 6th testcase)

Tiles: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Scores: [22, 22, 22, 23, 21, 24, 0, 22]

First score is 22, so take the highest tile in the middle <= 22, which is 22 itself.

Middle: [21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Stack: [22]

Remaining scores: [22, 22, 23, 21, 24, 0, 22] 

Next score is 22, so take the highest tile in the middle <= 22. Because 22 is already taken, the player has to take 21.

Middle: [23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Stack: [22, 21]

Remaining scores: [22, 23, 21, 24, 0, 22]

Next score is 22, but all numbers <= 22 are already taken. Therefore, the player loses the topmost tile on the stack (21), which is returned into the middle.

Middle: [21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Stack: [22]

Remaining scores: [23, 21, 24, 0, 22]

Next scores are 23, 21 and 24, so the player takes these tiles from the middle.

Middle: [25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Stack: [22, 23, 21, 24]

Remaining scores: [0, 22]

The player busts and scores zero. Therefore, the tile with the number 24 (topmost on the stack) is returned into the middle.

Middle: [24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Stack: [22, 23, 21]

Remaining scores: [22]

The last score is 22, but all tiles <= 22 are already taken, so the player loses the topmost tile on the stack (21).

Middle: [21, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Final Stack and Output: [22, 23]

Test cases

(with the topmost tile last in the output list)

Tiles: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Scores: [26, 30, 21]

Output: [26, 30, 21]



Tiles: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Scores: [35, 35, 36, 36]

Output: [35, 34, 36, 33]



Tiles: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Scores: [22, 17, 23, 19, 23]

Output: [23]



Tiles: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Scores: 

Output: 



Tiles: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Scores: [22, 17, 23, 19, 23, 0]

Output: 



Tiles: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36]

Scores: [22, 22, 22, 23, 21, 24, 0, 22]

Output: [22, 23]



Tiles: [1, 5, 9, 13, 17, 21, 26]

Scores: [6, 10, 23, 23, 23, 1, 0, 15]

Output: [5, 9, 21, 17, 13, 1]



Tiles: 

Scores: [4, 6, 1, 6]

Output: 

Sandbox

Haskell, 119 111 104 103 bytes

1 byte saved thanks to Ørjan Johansen

(#)=span.(<)

(a%(b:c))d|(g,e:h)<-b#d=(e:a)%c$g++h|g:h<-a,(i,j)<-g#d=h%c$i++g:j|1>0=a%c$d

(a%b)c=a

(%)

Try it online!

Assumes the tiles are sorted in descending order.

Not much fancy going on here. The first argument is the players pile, the second their scores and the third is the pile in the middle.

Japt, 24 bytes

Oof! That didn't work out as well as I thought it would!

Takes input in reverse order.

®=Va§Z)Ì?NpVjZ:VpNo)n

N¤

Try it

Charcoal, 35 bytes

Ｆη«≔⌈Φ講κιι¿ι«≔Φθ⁻κιθ⊞υι»¿υ⊞θ⊟υ»Ｉυ

Try it online! Link is to verbose version of code. Explanation:

Ｆη«

Loop over the scores.

≔⌈Φ講κιι

Look for the highest available tile.

¿ι«

If it exists then...

≔Φθ⁻κιθ

... remove the tile from the middle...

⊞υι

... and add it to the stack.

»¿υ

Otherwise, if the stack is not empty...

⊞θ⊟υ

Remove the latest tile from the stack and return it to the middle.

»Ｉυ

Print the resulting stack from oldest to newest.

Perl 6, 89 bytes

{my@x;@^a;@^b>>.&{@x=|@x,|(keys(@a∖@x).grep($_>=*).sort(-*)[0]//(try ~@x.pop&&()))};@x}

Try it online!

I think there's a few more bytes to be golfed off this...

JavaScript (ES6),  100 98  94 bytes

Takes input as (tiles)(scores). The tiles can be passed in any order.

t=>s=>s.map(x=>m[(g=_=>m[x]?x:g(x--))()?r.push(x)&&x:r.pop()]^=1,t.map(x=>m[x]=1,m=[r=]))&&r

Try it online!

Commented

t => s =>                 // t = tiles; s = scores

  s.map(x =>              // for each score x in s:

    m[                    //   update the 'middle':

      (g = _ =>           //     g = recursive function which decrements x until

        m[x] ? x : g(x--) //         m[x] is set and returns the final value

      )() ?               //     if x is not equal to 0 after that:

        r.push(x) && x    //       push x in the stack r and yield x

      :                   //     else:

        r.pop()           //       pop the last value from the stack

    ] ^= 1,               //   toggle the corresponding flag in m

    t.map(x =>            //   initialization of the 'middle': for each value x in t:

      m[x] = 1,           //     set m[x]

      m = [r = ]        //     the stack r is stored as the first entry of m,

                          //     which ensures that g will always stop when x = 0

    )                     //   end of initialization

  ) && r                  // end of main loop; return r