<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 sumQuery(vector<int>&preSum, int l, int r) 
    {
        if(l==0){return preSum[r];}
        return (preSum[r] - preSum[l-1]);
    }
    int stoneGameII(vector<int>& stones) 
    {
        int n = stones.size();
        
        vector<int> preSum(n, 0);
        preSum[0] = stones[0];
        for(int i=1; i<n; i++){preSum[i] = preSum[i-1] + stones[i];}
        
        vector<vector<int>> maxStones(n, vector<int>(2*n + 3, 0));
        vector<vector<int>> optimalToTake(n, vector<int>(2*n + 3, 0));
        
        for(int i=n-1; i>=0; i--) {
            int n_curr = (n - i);
            for(int m=2*n; m>=1; m--) {
                for(int choose=1; choose<=min(2*m, n_curr); choose++) {
                    
                    int max_val = 0;
                    int curr_sum_chosen = sumQuery(preSum, i, i+choose-1);
                    
                    int i_alice_1 = i;
                    int i_bob = i + choose;
                    int i_alice_2;
                    
                    if(i_bob >= n) {
                        max_val = curr_sum_chosen;
                    }
                    else {
                        i_alice_2 = i_bob + optimalToTake[i_bob][max(m, choose)];
                        if(i_alice_2 >= n) {
                            max_val = curr_sum_chosen;
                        } 
                        else {
                            int m_new = max(max(m, choose), optimalToTake[i_bob][max(m, choose)]);
                            max_val = curr_sum_chosen + maxStones[i_alice_2][m_new];
                        }
                    }
                    
                    if(max_val >= maxStones[i][m]) {
                        maxStones[i][m] = max_val;
                        optimalToTake[i][m] = choose;
                    }
                }
            }
        }
        
        return maxStones[0][1];
    }
};
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