Minimum Cost Tree From Leaf Values
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1130.Minimum%20Cost%20Tree%20From%20Leaf%20Values/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1130.Minimum%20Cost%20Tree%20From%20Leaf%20Values/images/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>
Stack; Greedy; Array; Dynamic Programming; Monotonic Stack
class Solution:
    def mctFromLeafValues(self, arr: List[int]) -> int:
        @cache
        def dfs(i: int, j: int) -> Tuple:
            if i == j:
                return 0, arr[i]
            s, mx = inf, -1
            for k in range(i, j):
                s1, mx1 = dfs(i, k)
                s2, mx2 = dfs(k + 1, j)
                t = s1 + s2 + mx1 * mx2
                if s > t:
                    s = t
                    mx = max(mx1, mx2)
            return s, mx

        return dfs(0, len(arr) - 1)[0]
