<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>

class Solution {
public:
    //You can easily optimise the below code in terms of space. Take only 2 rows.
    //Logic: https://www.youtube.com/watch?v=QZ9edJ0JCPw&list=PLDzeHZWIZsTryvtXdMr6rPh4IDexB5NIA&index=130

    //for each possible way of choosing k/3 slices without choosing adjacent slices, there is a particular
    //order for picking up those slices according to the rules of this question.
    int maxSizeSlices(vector<int>& slices) {
        int k=slices.size();

        vector<vector<int>> dp(k/3+1, vector<int>(k, 0));
        dp[1][0]=slices[0];
        for(int i=1; i<k-1; i++){
            dp[1][i]=max(slices[i], dp[1][i-1]);
        }

        for(int i=2; i<=k/3; i++){
            for(int j=2; j<k-1; j++){
                dp[i][j] = max(slices[j] + dp[i-1][j-2], dp[i][j-1]);
            }
        }
        int ans1=dp[k/3][k-2];

        for(int i=0; i<=k/3; i++){
            for(int j=0; j<k; j++){
                dp[i][j]=0;
            }
        }
        dp[1][1]=slices[1];
        for(int i=2; i<k; i++){
            dp[1][i]=max(slices[i], dp[1][i-1]);
        }

        for(int i=2; i<=k/3; i++){
            for(int j=3; j<k; j++){
                dp[i][j] = max(slices[j] + dp[i-1][j-2], dp[i][j-1]);
            }
        }
        int ans2=dp[k/3][k-1];
        return max(ans1, ans2);
    }
};
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