<p>Alice and Bob continue their games with piles of stones. There are a number of piles <strong>arranged in a row</strong>, and each pile has a positive integer number of stones <code>piles[i]</code>. The objective of the game is to end with the most stones.</p>

<p>Alice and Bob take turns, with Alice starting first.</p>

<p>On each player&#39;s turn, that player can take <strong>all the stones</strong> in the <strong>first</strong> <code>X</code> remaining piles, where <code>1 &lt;= X &lt;= 2M</code>. Then, we set <code>M = max(M, X)</code>. Initially, M = 1.</p>

<p>The game continues until all the stones have been taken.</p>

<p>Assuming Alice and Bob play optimally, return the maximum number of stones Alice can get.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">piles = [2,7,9,4,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">10</span></p>

<p><strong>Explanation:</strong></p>

<ul>
	<li>If Alice takes one pile at the beginning, Bob takes two piles, then Alice takes 2 piles again. Alice can get <code>2 + 4 + 4 = 10</code> stones in total.</li>
	<li>If Alice takes two piles at the beginning, then Bob can take all three piles left. In this case, Alice get <code>2 + 7 = 9</code> stones in total.</li>
</ul>

<p>So we return 10 since it&#39;s larger.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">piles = [1,2,3,4,5,100]</span></p>

<p><strong>Output:</strong> <span class="example-io">104</span></p>
</div>

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

<ul>
	<li><code>1 &lt;= piles.length &lt;= 100</code></li>
	<li><code>1 &lt;= piles[i]&nbsp;&lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    // dp is a 3D array used for memoization to store results of subproblems
    vector<vector<vector<int>>> dp;
    
    // Helper function to calculate the maximum score Alice can achieve
    int score(bool alice, int idx, int M, vector<int>& piles) {
        // Base case: if we've reached the end of the piles, no more stones can be taken
        if (idx >= piles.size()) {
            return 0;
        }
        
        // Check if the result for this state is already computed
        if (dp[alice][idx][M] != -1) {
            return dp[alice][idx][M];
        }
        
        // Initialize the result based on whose turn it is
        int res = alice ? 0 : INT_MAX;
        int total = 0; // Total number of stones taken in this turn
        
        // Consider taking x piles where x ranges from 1 to 2 * M
        for (int x = 1; x <= 2 * M; ++x) {
            if (idx + x > piles.size()) { // If taking x piles exceeds available piles, stop
                break;
            }
            total += piles[idx + x - 1]; // Add stones from the current pile to total
            
            // Recursively calculate the score for the next state
            if (alice) {
                res = max(res, total + score(!alice, idx + x, max(M, x), piles));
            } else {
                res = min(res, score(!alice, idx + x, max(M, x), piles));
            }
        }
        
        // Store the result in the memoization array
        dp[alice][idx][M] = res;
        return res;
    }
    
    int stoneGameII(vector<int>& piles) {
        int n = piles.size();
        // Initialize the memoization array with -1
        dp = vector<vector<vector<int>>>(2, vector<vector<int>>(n, vector<int>(n + 1, -1)));
        
        // Start the game with Alice's turn, starting at index 0, with M initially set to 1
        return score(true, 0, 1, piles);
    }
};
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