<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>

//Recursion + Memo
//T.C : O(n^3)
//S.C : O(n^3)
class Solution {
public:
    
    int n;
    int t[2][101][101]; 
    int solveForAlice(vector<int>& piles, int person, int i, int M) { //person = 1/0 = alice/bob
        
        if(i >= n)
            return 0;
        
        if(t[person][i][M] != -1)
            return t[person][i][M];
        
        int result = (person == 1) ? -1 : INT_MAX;
        
        int stones = 0;
        
        for(int x = 1; x <= min(2*M, n-i); x++) {
            
            stones += piles[i+x-1];
            
            if(person == 1) { //Alice
                result = max(result, stones + solveForAlice(piles, 0, i+x, max(M, x)));
            } else { //Bob
                result = min(result, solveForAlice(piles, 1, i+x, max(M, x)));
            }
            
        }
        
        return t[person][i][M] = result;
        
    }
    
    int stoneGameII(vector<int>& piles) {
        n = piles.size();
        
        memset(t, -1, sizeof(t));
        
        return solveForAlice(piles, 1, 0, 1); 
    }
};
//Alice = 1, Bob = 0
10371