<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:
    bool stoneGame(vector<int>& v) {
        int n = v.size();
        // dp[l][r][k] stores whether the current player can win when considering subarray [l, r]
        // k = 0 -> Player 1's turn, k = 1 -> Player 2's turn
        vector<vector<vector<int>>> dp(n, vector<vector<int>>(n, vector<int>(2, 0)));

        // Base case: when l == r, only one element is available
        for (int i = 0; i < n; i++) {
            dp[i][i][0] = v[i]; // Player 1 picks the only element
            dp[i][i][1] = 0; // Player 2 doesn't get a chance to pick when there's only one element
        }

        // Iterate over subarray lengths
        for (int length = 2; length <= n; length++) {
            for (int l = 0; l <= n - length; l++) {
                int r = l + length - 1;

                // Calculate for Player 1 (k = 0)
                int pickLeft = v[l] + dp[l + 1][r][1];
                int pickRight = v[r] + dp[l][r - 1][1];
                dp[l][r][0] = max(pickLeft, pickRight);

                // Calculate for Player 2 (k = 1)
                pickLeft = dp[l + 1][r][0];
                pickRight = dp[l][r - 1][0];
                dp[l][r][1] = min(pickLeft, pickRight);
            }
        }

        // The result is based on whether Player 1 can win considering the whole array
        return dp[0][n - 1][0] > dp[0][n - 1][1];
    }
};

84888