<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>

#include <vector>
#include <string>

class Solution {
public:
    // Recursive function to calculate the maximum score
    int calculate(vector<int>& stoneValues, vector<int>& dynamicProgramming, int currentIndex, int totalStones) {
        // Base case: if the current index is beyond the last stone, return 0
        if (currentIndex >= totalStones)
            return 0;

        // If the value for the current index is already calculated, return it
        if (dynamicProgramming[currentIndex] != -1)
            return dynamicProgramming[currentIndex];

        // Initialize the result for the current index
        int result = INT_MIN;

        // Calculate the score by considering the current stone alone
        result = max(result, stoneValues[currentIndex] - calculate(stoneValues, dynamicProgramming, currentIndex + 1, totalStones));

        // Calculate the score by considering the current stone and the next one
        if (currentIndex + 1 < totalStones)
            result = max(result, stoneValues[currentIndex] + stoneValues[currentIndex + 1] - calculate(stoneValues, dynamicProgramming, currentIndex + 2, totalStones));

        // Calculate the score by considering the current stone and the next two stones
        if (currentIndex + 2 < totalStones)
            result = max(result, stoneValues[currentIndex] + stoneValues[currentIndex + 1] + stoneValues[currentIndex + 2] - calculate(stoneValues, dynamicProgramming, currentIndex + 3, totalStones));

        // Memoize the result and return it
        dynamicProgramming[currentIndex] = result;
        return dynamicProgramming[currentIndex];
    }

    // Function to determine the winner of the stone game
    string stoneGameIII(vector<int>& stoneValues) {
        int totalStones = stoneValues.size();
        // Initialize a dynamic programming array with -1
        vector<int> dynamicProgramming(totalStones + 1, -1);
        // Calculate the score for the starting position
        int winnerScore = calculate(stoneValues, dynamicProgramming, 0, totalStones);

        // Determine the winner based on the calculated score
        if (winnerScore > 0)
            return "Alice";
        else if (winnerScore < 0)
            return "Bob";
        else
            return "Tie";
    }
};
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