<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>

class Solution {
public:

    pair<int, int> inner(const vector<int>& piles, int l, int r, bool is_first, unordered_map<int, unordered_map<int, pair<int, int>>>& dp) {
        if (l > r) return {0, 0};
        if (dp.find(l)!=dp.end() and dp[l].find(r)!=dp[l].end()) return dp[l][r];

        pair<int, int> take_left = inner(piles, l+1, r, !is_first, dp);
        pair<int, int> take_right = inner(piles, l, r-1, !is_first, dp);
        
        pair<int, int> res;
        if (is_first) {
            if (take_left.first + piles[l] >= take_right.first + piles[r]) res = {take_left.first + piles[l], take_left.second};
            else res = {take_right.first+piles[r], take_right.second};
        }
        else {
            if (take_left.second + piles[l] >= take_right.second + piles[r]) res = {take_left.first, take_left.second+piles[l]};
            else res = {take_right.first, take_right.second+piles[r]};
        }
        dp[l][r] = res;
        return res;
    }

    bool stoneGame(vector<int>& piles) {
        // play optimally means maximizing the number of stones at each time step    
        unordered_map<int, unordered_map<int, pair<int, int>>> dp;
        pair<int, int> game_res = inner(piles, 0, piles.size()-1, true, dp);
        if (game_res.first >= game_res.second) return true;
        else return false;
    }
};
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