<p>Alice and Bob play a game with piles of stones. There are an <strong>even</strong> number of piles arranged in a row, and each pile has a <strong>positive</strong> integer number of stones <code>piles[i]</code>.</p>

<p>The objective of the game is to end with the most stones. The <strong>total</strong> number of stones across all the piles is <strong>odd</strong>, so there are no ties.</p>

<p>Alice and Bob take turns, with <strong>Alice starting first</strong>. Each turn, a player takes the entire pile of stones either from the <strong>beginning</strong> or from the <strong>end</strong> of the row. This continues until there are no more piles left, at which point the person with the <strong>most stones wins</strong>.</p>

<p>Assuming Alice and Bob play optimally, return <code>true</code><em> if Alice wins the game, or </em><code>false</code><em> if Bob wins</em>.</p>

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

<pre>
<strong>Input:</strong> piles = [5,3,4,5]
<strong>Output:</strong> true
<strong>Explanation:</strong> 
Alice starts first, and can only take the first 5 or the last 5.
Say she takes the first 5, so that the row becomes [3, 4, 5].
If Bob takes 3, then the board is [4, 5], and Alice takes 5 to win with 10 points.
If Bob takes the last 5, then the board is [3, 4], and Alice takes 4 to win with 9 points.
This demonstrated that taking the first 5 was a winning move for Alice, so we return true.
</pre>

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

<pre>
<strong>Input:</strong> piles = [3,7,2,3]
<strong>Output:</strong> true
</pre>

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

<ul>
	<li><code>2 &lt;= piles.length &lt;= 500</code></li>
	<li><code>piles.length</code> is <strong>even</strong>.</li>
	<li><code>1 &lt;= piles[i] &lt;= 500</code></li>
	<li><code>sum(piles[i])</code> is <strong>odd</strong>.</li>
</ul>

unordered_map<int, int> memo;
unordered_map<int, bool> found;
int nums[500];
int solve(int a, int b){
    if(found[a + b * SHRT_MAX]) return memo[a + b * SHRT_MAX];
    found[a + b * SHRT_MAX] = true;
    return memo[a + b * SHRT_MAX] = max(nums[a] - solve(a + 1, b), nums[b] - solve(a, b - 1));
}
class Solution {
public:
    bool stoneGame(vector<int>& piles) {
        memo.clear(), found.clear();
        for(int i = 0, n = piles.size(); i < n; i++) nums[i] = piles[i], memo[i + i * SHRT_MAX] = i, found[i + i * SHRT_MAX] = true;
        return solve(0, piles.size() - 1);
    }
};
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