<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 {
    vector<unordered_map<int, std::optional<int>>> memo;
public:
    int stoneGameII(vector<int>& piles) {
        // dynamic programming, two dimensions
        deque<int> prefix_sum;
        std::partial_sum(piles.rbegin(), piles.rend(), std::front_inserter(prefix_sum));
        prefix_sum.push_back(0);

        memo.resize(piles.size()+1);

        return dp(piles, piles.size(), prefix_sum, 0, 1);
    }

    int dp(vector<int>&piles, int n, deque<int> &prefix_sum, int idx, int M) {
        if (idx == n)
            return 0;
        
        if (memo[idx][M]) {
            return *memo[idx][M];
        }
        
        int sum = 0, ans = 0;
        for(int i = idx; i < std::min(n, idx+2*M); i++) {
            sum += piles[i];
            int temp = dp(piles, n, prefix_sum, i+1, std::max(M, (i-idx+1)));
            // assuming the next player plays optimally starting from i+1
            int complement = prefix_sum[i+1] - temp;
            int candidate = sum + complement;

            ans = std::max(ans, candidate);
        }

        memo[idx][M] = ans;
        return ans;
    }
};
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