<p>There is a pizza with <code>3n</code> slices of varying size, you and your friends will take slices of pizza as follows:</p>

<ul>
	<li>You will pick <strong>any</strong> pizza slice.</li>
	<li>Your friend Alice will pick the next slice in the anti-clockwise direction of your pick.</li>
	<li>Your friend Bob will pick the next slice in the clockwise direction of your pick.</li>
	<li>Repeat until there are no more slices of pizzas.</li>
</ul>

<p>Given an integer array <code>slices</code> that represent the sizes of the pizza slices in a clockwise direction, return <em>the maximum possible sum of slice sizes that you can pick</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/18/sample_3_1723.png" style="width: 500px; height: 266px;" />
<pre>
<strong>Input:</strong> slices = [1,2,3,4,5,6]
<strong>Output:</strong> 10
<strong>Explanation:</strong> Pick pizza slice of size 4, Alice and Bob will pick slices with size 3 and 5 respectively. Then Pick slices with size 6, finally Alice and Bob will pick slice of size 2 and 1 respectively. Total = 4 + 6.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/02/18/sample_4_1723.png" style="width: 500px; height: 299px;" />
<pre>
<strong>Input:</strong> slices = [8,9,8,6,1,1]
<strong>Output:</strong> 16
<strong>Explanation:</strong> Pick pizza slice of size 8 in each turn. If you pick slice with size 9 your partners will pick slices of size 8.
</pre>

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

<ul>
	<li><code>3 * n == slices.length</code></li>
	<li><code>1 &lt;= slices.length &lt;= 500</code></li>
	<li><code>1 &lt;= slices[i] &lt;= 1000</code></li>
</ul>

static const int speedup = [] static { ios::sync_with_stdio(0); cin.tie(0); return 0;}();
constexpr int MSZ = 167;
int buf1[MSZ], buf2[MSZ];

class Solution {
public:
    int maxSizeSlices(vector<int>& slices) {
        int N = size(slices), n = N/3, *p1 = buf1, *p2 = buf2;

        // Take N-1 (so not 0)
        fill(p2+1, p2+n+1, slices[N-1]); // N-1
        fill(p1+1, p1+n+1, max(slices[N-1], slices[N-2])); // N-2
        for (int i = N-3; i > 0; --i) {
            for (int k = n, v = slices[i]; k; --k) p2[k] = max(v + p2[k-1], p1[k]);
            swap(p1, p2);
        }
        int s1 = p1[n];

        // Don't take N-1 (can take 0)
        fill(p2+1, p2+n+1, 0); // N-1
        fill(p1+1, p1+n+1, slices[N-2]); // N-2
        for (int i = N-3; i >= 0; --i) {
            for (int k = n, v = slices[i]; k; --k) p2[k] = max(v + p2[k-1], p1[k]);
            swap(p1, p2);
        }
        return max(s1, p1[n]);
    }
};
10500