<p>You are playing a variation of the game Zuma.</p>

<p>In this variation of Zuma, there is a <strong>single row</strong> of colored balls on a board, where each ball can be colored red <code>&#39;R&#39;</code>, yellow <code>&#39;Y&#39;</code>, blue <code>&#39;B&#39;</code>, green <code>&#39;G&#39;</code>, or white <code>&#39;W&#39;</code>. You also have several colored balls in your hand.</p>

<p>Your goal is to <strong>clear all</strong> of the balls from the board. On each turn:</p>

<ul>
	<li>Pick <strong>any</strong> ball from your hand and insert it in between two balls in the row or on either end of the row.</li>
	<li>If there is a group of <strong>three or more consecutive balls</strong> of the <strong>same color</strong>, remove the group of balls from the board.
	<ul>
		<li>If this removal causes more groups of three or more of the same color to form, then continue removing each group until there are none left.</li>
	</ul>
	</li>
	<li>If there are no more balls on the board, then you win the game.</li>
	<li>Repeat this process until you either win or do not have any more balls in your hand.</li>
</ul>

<p>Given a string <code>board</code>, representing the row of balls on the board, and a string <code>hand</code>, representing the balls in your hand, return <em>the <strong>minimum</strong> number of balls you have to insert to clear all the balls from the board. If you cannot clear all the balls from the board using the balls in your hand, return </em><code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> board = &quot;WRRBBW&quot;, hand = &quot;RB&quot;
<strong>Output:</strong> -1
<strong>Explanation:</strong> It is impossible to clear all the balls. The best you can do is:
- Insert &#39;R&#39; so the board becomes WRR<u>R</u>BBW. W<u>RRR</u>BBW -&gt; WBBW.
- Insert &#39;B&#39; so the board becomes WBB<u>B</u>W. W<u>BBB</u>W -&gt; WW.
There are still balls remaining on the board, and you are out of balls to insert.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> board = &quot;WWRRBBWW&quot;, hand = &quot;WRBRW&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> To make the board empty:
- Insert &#39;R&#39; so the board becomes WWRR<u>R</u>BBWW. WW<u>RRR</u>BBWW -&gt; WWBBWW.
- Insert &#39;B&#39; so the board becomes WWBB<u>B</u>WW. WW<u>BBB</u>WW -&gt; <u>WWWW</u> -&gt; empty.
2 balls from your hand were needed to clear the board.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> board = &quot;G&quot;, hand = &quot;GGGGG&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> To make the board empty:
- Insert &#39;G&#39; so the board becomes G<u>G</u>.
- Insert &#39;G&#39; so the board becomes GG<u>G</u>. <u>GGG</u> -&gt; empty.
2 balls from your hand were needed to clear the board.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= board.length &lt;= 16</code></li>
	<li><code>1 &lt;= hand.length &lt;= 5</code></li>
	<li><code>board</code> and <code>hand</code> consist of the characters <code>&#39;R&#39;</code>, <code>&#39;Y&#39;</code>, <code>&#39;B&#39;</code>, <code>&#39;G&#39;</code>, and <code>&#39;W&#39;</code>.</li>
	<li>The initial row of balls on the board will <strong>not</strong> have any groups of three or more consecutive balls of the same color.</li>
</ul>

class Solution:
    def findMinStep(self, board: str, hand: str) -> int:
        # Sort the hand to facilitate pruning
        hand = ''.join(sorted(hand))
        
        @cache
        def fn(board, hand):
            """Return the minimum number of balls to insert to clear the board."""
            # Base case: if the board is empty, no more balls are needed
            if not board:
                return 0
            # If the hand is empty and the board is not, it's impossible to clear the board
            if not hand:
                return float('inf')
            
            ans = float('inf')
            # Iterate over each ball in the hand
            for i, ch in enumerate(hand):
                # Pruning: only consider the first occurrence of each color in the hand
                if i == 0 or hand[i-1] != ch:
                    # Create a new hand without the current ball
                    hh = hand[:i] + hand[i+1:]
                    # Try inserting the ball at every possible position in the board
                    for j in range(0, len(board)):
                        # Pruning: only insert if it matches the current or previous ball
                        if ch == board[j] or (j > 0 and board[j-1] == board[j]):
                            # Insert the ball and attempt to clear the board
                            bb, nn = "", board[:j] + ch + board[j:]
                            # Continuously remove groups of 3 or more balls
                            while bb != nn:
                                bb, nn = nn, ""
                                for k, grp in groupby(bb):
                                    x = len(list(grp))
                                    # Only keep groups of less than 3 balls
                                    if x < 3:
                                        nn += k * x
                            # Recur with the new board and hand, and update the answer
                            ans = min(ans, 1 + fn(bb, hh))
            return ans
        
        # Calculate the result and return -1 if it's impossible to clear the board
        result = fn(board, hand)
        return result if result < float('inf') else -1
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