<p>Given an array <code>arr</code> of positive integers, consider all binary trees such that:</p>

<ul>
	<li>Each node has either <code>0</code> or <code>2</code> children;</li>
	<li>The values of <code>arr</code> correspond to the values of each <strong>leaf</strong> in an in-order traversal of the tree.</li>
	<li>The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree, respectively.</li>
</ul>

<p>Among all possible binary trees considered, return <em>the smallest possible sum of the values of each non-leaf node</em>. It is guaranteed this sum fits into a <strong>32-bit</strong> integer.</p>

<p>A node is a <strong>leaf</strong> if and only if it has zero children.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/10/tree1.jpg" style="width: 500px; height: 169px;" />
<pre>
<strong>Input:</strong> arr = [6,2,4]
<strong>Output:</strong> 32
<strong>Explanation:</strong> There are two possible trees shown.
The first has a non-leaf node sum 36, and the second has non-leaf node sum 32.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/08/10/tree2.jpg" style="width: 224px; height: 145px;" />
<pre>
<strong>Input:</strong> arr = [4,11]
<strong>Output:</strong> 44
</pre>

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

<ul>
	<li><code>2 &lt;= arr.length &lt;= 40</code></li>
	<li><code>1 &lt;= arr[i] &lt;= 15</code></li>
	<li>It is guaranteed that the answer fits into a <strong>32-bit</strong> signed integer (i.e., it is less than 2<sup>31</sup>).</li>
</ul>

class Solution {
public:
    int mctFromLeafValues(vector<int>& a) {
        int n = a.size();

        vector<vector<int>>mx(n, vector<int>(n));
        vector<vector<int>>dp(n, vector<int>(n, INT_MAX));
        for(int i = 0; i<n; i++)
        {
            dp[i][i] = 0;
            int val = a[i];
            for(int j = i; j<n; j++)
            {
                val = max(val, a[j]);
                mx[i][j] = val;
            }
        }

        for(int len = 2; len<=n; len++)
        {
            for(int i = 0; i+len-1<n; i++)
            {
                for(int j = i+1; j<=i+len-1; j++)
                {
                    int left = dp[i][j-1], right = dp[j][i+len-1];
                    dp[i][i+len-1] = min(dp[i][i+len-1], left+right+mx[i][j-1]*mx[j][i+len-1]);
                }
            }
        }
        return dp[0][n-1];
    }
};
11000