<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:
    int stoneGameII(vector<int>& piles) {
        ios_base::sync_with_stdio(false);
        cout.tie(0);
        int n = piles.size();

        vector<vector<vector<int>>> dp(2, vector<vector<int>>(n + 1, vector<int>(n + 1, 0)));

        vector<int> suffixSum(n + 1, 0);

        for (int i = n - 1; i >= 0; --i) {
            suffixSum[i] = piles[i] + suffixSum[i + 1];
        }

    
        for (int index = n - 1; index >= 0; --index) {
            for (int M = 1; M <= n; ++M) {
                for (int turn = 0; turn < 2; ++turn) {
                    if (!turn) { 
                        int maxScore = INT_MIN;
                        for (int i = 0; i < min(2*M,n-index); ++i) {
                            int currentSum = suffixSum[index] - suffixSum[index + i + 1];
                        maxScore = max(maxScore, currentSum + dp[1][index + i + 1][max(M, i + 1)]);
                        }
                        dp[0][index][M] = maxScore;
                    } else { 
                        int minScore = INT_MAX;
                        for (int i = 0; i < 2 * M && index + i < n; ++i) {
                            int opponentScore = dp[0][index + i + 1][max(M, i + 1)];
                            minScore = min(minScore, opponentScore);
                        }
                        dp[1][index][M] = minScore;
                    }
                }
            }
        }

        return dp[0][0][1];
    }
};

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