<p>Alice and Bob continue their games with piles of stones. There are several stones <strong>arranged in a row</strong>, and each stone has an associated value which is an integer given in the array <code>stoneValue</code>.</p>

<p>Alice and Bob take turns, with Alice starting first. On each player&#39;s turn, that player can take <code>1</code>, <code>2</code>, or <code>3</code> stones from the <strong>first</strong> remaining stones in the row.</p>

<p>The score of each player is the sum of the values of the stones taken. The score of each player is <code>0</code> initially.</p>

<p>The objective of the game is to end with the highest score, and the winner is the player with the highest score and there could be a tie. The game continues until all the stones have been taken.</p>

<p>Assume Alice and Bob <strong>play optimally</strong>.</p>

<p>Return <code>&quot;Alice&quot;</code><em> if Alice will win, </em><code>&quot;Bob&quot;</code><em> if Bob will win, or </em><code>&quot;Tie&quot;</code><em> if they will end the game with the same score</em>.</p>

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

<pre>
<strong>Input:</strong> stoneValue = [1,2,3,7]
<strong>Output:</strong> &quot;Bob&quot;
<strong>Explanation:</strong> Alice will always lose. Her best move will be to take three piles and the score become 6. Now the score of Bob is 7 and Bob wins.
</pre>

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

<pre>
<strong>Input:</strong> stoneValue = [1,2,3,-9]
<strong>Output:</strong> &quot;Alice&quot;
<strong>Explanation:</strong> Alice must choose all the three piles at the first move to win and leave Bob with negative score.
If Alice chooses one pile her score will be 1 and the next move Bob&#39;s score becomes 5. In the next move, Alice will take the pile with value = -9 and lose.
If Alice chooses two piles her score will be 3 and the next move Bob&#39;s score becomes 3. In the next move, Alice will take the pile with value = -9 and also lose.
Remember that both play optimally so here Alice will choose the scenario that makes her win.
</pre>

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

<pre>
<strong>Input:</strong> stoneValue = [1,2,3,6]
<strong>Output:</strong> &quot;Tie&quot;
<strong>Explanation:</strong> Alice cannot win this game. She can end the game in a draw if she decided to choose all the first three piles, otherwise she will lose.
</pre>

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

<ul>
	<li><code>1 &lt;= stoneValue.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-1000 &lt;= stoneValue[i] &lt;= 1000</code></li>
</ul>

class Solution:
    def stoneGameIII(self, stoneValue: List[int]) -> str:
        total_score = sum(stoneValue)
        cache = {}
        alice_score = self.game(stoneValue, 0, len(stoneValue)-1, cache)
        bob_score = total_score - alice_score
        
        if alice_score > bob_score:
            return "Alice"
        elif alice_score < bob_score:
            return "Bob"
        else:
            return "Tie"

    def game(self, stoneValue, low, high, cache):
        if low > high:
            return 0
        if (low, high) in cache:
            return cache[(low, high)]

        # Wow, there are quite a few probabilities...
        planA = stoneValue[low] + min(
            self.game(stoneValue, low+2, high, cache),
            self.game(stoneValue, low+3, high, cache),
            self.game(stoneValue, low+4, high, cache)
        )
        if low + 1 > high:
            planB = float("-inf")
        else:
            planB = stoneValue[low] + stoneValue[low+1] + min(
                self.game(stoneValue, low+3, high, cache),
                self.game(stoneValue, low+4, high, cache),
                self.game(stoneValue, low+5, high, cache)
            )
        if low + 2 > high:
            planC = float("-inf")
        else:
            planC = stoneValue[low] + stoneValue[low+1] + stoneValue[low+2] + min(
                self.game(stoneValue, low+4, high, cache),
                self.game(stoneValue, low+5, high, cache),
                self.game(stoneValue, low+6, high, cache)
            )
        # print(f"Now low is {low}, and the 3 plans are {planA}, {planB}, {planC}")
        output = max(planA, planB, planC)
        cache[(low, high)] = output
        return output




        
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