<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>

from functools import lru_cache
from math import prod

class Solution:
    def mctFromLeafValues(self, arr: List[int]) -> int:
        self.arr = arr
        self.l = len(arr)
        return self.mctFromLeafValuesRecu(0,self.l-1)

    @lru_cache(maxsize=1024)
    def mctFromLeafValuesRecu(self,st,en):
        if en-st+1==2:
            return self.arr[st]*self.arr[en]
        elif en-st+1==3:
            op1 = self.arr[st]*max(self.arr[st+1:en+1]) + prod(self.arr[st+1:en+1])
            op2 = self.arr[en]*max(self.arr[st:en]) + prod(self.arr[st:en])
            return min(op1,op2)
        else:
            sep = list(range(st+1,en+1))
            op = 40**9
            for s in sep:
                if s==st+1:
                    r = self.arr[st]*max(self.arr[s:en+1]) + self.mctFromLeafValuesRecu(st+1,en)
                elif s==en:
                    r = self.arr[en]*max(self.arr[st:s]) + self.mctFromLeafValuesRecu(st,en-1)
                else:
                    r = max(self.arr[st:s])*max(self.arr[s:en+1]) + self.mctFromLeafValuesRecu(st,s-1) + self.mctFromLeafValuesRecu(s,en)
                op = min(op, r)
            return op
            
16900