<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>

from functools import cache
class Solution:
    """
    each time take 1 or 2 or 3 stones, alice starts first

    dp(i): optimal score and optimal move, for stone[i...n-1] for current player
    dp(i) = max,argmax(stone[i] + dp(i+1+dp(i+1).move).score,
            stone[i]+stone[i+1]+ dp(i+2+dp(i+2).move).score,
            stone[i]+stone[i+1]+stone[i+2]+dp(i+3 + dp(i+3).move).score
    if tie in argmax, take either one
    base case: if dp(>=n), score=0, move=0
    dp(0).score is alice's score, compare dp(0).score with sum(stone)-dp(0).score

    time and space O(n)
    """
    def stoneGameIII(self, stoneValue: List[int]) -> str:
        n = len(stoneValue)
        # (score, move)
        @cache
        def dp(i):
            if i >= n: return 0, 0
            a = stoneValue[i] + dp(i+1 + dp(i+1)[1])[0]
            score, move = a, 1
            if i <= n-2:
                b = stoneValue[i] + stoneValue[i+1] + dp(i+2 + dp(i+2)[1])[0]
                if b > score: score, move = b, 2
            if i <= n-3:
                c = stoneValue[i] + stoneValue[i+1] + stoneValue[i+2] + dp(i+3+dp(i+3)[1])[0]
                if c > score: score, move = c, 3
            return score, move

        alice = dp(0)[0]
        bob = sum(stoneValue) - alice
        if alice > bob: return "Alice"
        elif alice < bob: return "Bob"
        return "Tie"
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