<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 {
private:
  typedef vector<pair<bool, int>> Cache;
public:
  string stoneGameIII(const vector<int>& values) {
    Cache cache(2 * values.size());

    const int alice_score = score(values, 0, true, cache);

    const int total = accumulate(
      values.begin(),
      values.end(),
      0,
      [](int r, int v) { return r + v; }
    );

    const int bob_score = total - alice_score;

    if (alice_score > bob_score) {
      return "Alice";
    }

    if (bob_score > alice_score) {
      return "Bob";
    }

    return "Tie";
  }

  int score(const vector<int>& values, int i, bool alice_turn, Cache& cache) {
    if (i >= values.size()) return 0;

    auto& cached = cache[(i << 1) + alice_turn];
    if (cached.first) return cached.second;

    const int last = min(3, (int) values.size() - i);

    if (alice_turn) {
      int m = INT_MIN;
      for (int x = 1; x <= last; ++x) {
        const int s = accumulate(
          values.begin() + i,
          values.begin() + i + x,
          0,
          [](int r, int v) { return r + v; }
        );

        m = max(m, s + score(values, i+x, false, cache));
      }

      cached.first = true;
      return cached.second = m;
    }

    int m = INT_MAX;
    for (int x = 1; x <= last; ++x) {
      m = min(m, score(values, i+x, true, cache));
    }
    cached.first = true;
    return cached.second = m;
  }
};

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