<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 maxX(int move) {
        // largest value of X available before move # (currently an approximation)
        // closer approximation probably exists, would be more efficient
        return move + 1;
    }
    int stoneGameII(std::vector<int>& piles) {
        int size = piles.size();
        int last = size - 1;

        // how many stones are left on the board before move # (indexing from 0)
        std::vector<int> remaining;
        int p = 0;
        for(int i : piles) {
            p += i;
        }
        for(int i = last; i >= 0; i--) {
            remaining.push_back(p);
            p -= piles[last - i];
        }

        // create a field for all the available options of moves left and X-1
        // the value indicates how many stones can the person playing from that position get
        std::vector<std::vector<int>> moves; 
        for(int i = 0; i < size; i++) {
            moves.push_back(std::vector<int>(maxX(i)));
        }
        // add zero to the end - after the last move there are no points to gain
        moves.push_back(std::vector<int>(maxX(size), 0)); 


        for(int i = last; i >= 0; i--) {
            for(int j = 0; j < maxX(i); j++) {
                // finding the min value the opposing player can get
                int min = 0x7fffffff;
                for(int k = 0; k < (size - i < 2 * j + 2 ? size - i : 2 * j + 2); k++) {
                    int value = moves[i + k + 1][k > j ? k : j];
                    min = value < min ? value : min;
                }
                // the player starting gets all the other points
                moves[i][j] = remaining[i] - min;
            }
        }
        // most points from position at move 0 with X = 1 (starting position)
        return moves[0][0];
    }
};
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