<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:
    def mctFromLeafValues(self, arr: List[int]) -> int:
        # define a recurrence relation solve(i, j) where i j are the bounds of the current subtree
        # split the subtree into 2 repeatedly and get the min sum of products
        
        dp = [[0 for i in range(len(arr))] for j in range(len(arr))]

        def solve(i, j):
            if i == j:
                return 0

            if dp[i][j] > 0:
                return dp[i][j]

            ans = float('inf')
            for x in range(i, j):
                left = solve(i, x)
                right = solve(x + 1, j)
                max_left = max(arr[i:x + 1])
                max_right = max(arr[x + 1:j + 1])

                ans = min(ans, left + right + max_left * max_right)

            dp[i][j] = ans
            return dp[i][j]

        return solve(0, len(arr) - 1)
17200