human
I found an interesting problem on https://leetcode.com/problems/minimum-cost-tree-from-leaf-values/:
Given an array arr of positive integers, consider all binary trees such that:

Each node has either 0 or 2 children;
The values of arr correspond to the values of each leaf in an in-order traversal of the tree.  (Recall that a node is a leaf if and only if it has 0 children.)
The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree respectively.

Among all possible binary trees considered, return the smallest possible sum of the values of each non-leaf node.  It is guaranteed this sum fits into a 32-bit integer.
 
Example 1:
Input: arr = [6,2,4]
Output: 32
Explanation:
There are two possible trees.  The first has non-leaf node sum 36, and the second has non-leaf node sum 32.

    24            24
   /  \          /  \
  12   4        6    8
 /  \               / \
6    2             2   4

 
Constraints:

2 <= arr.length <= 40
1 <= arr[i] <= 15
It is guaranteed that the answer fits into a 32-bit signed integer (ie. it is less than 2^31).

I tried it in Python, but could not do it. Can you solve it?

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Give me a the Python program you requested?
assistant
#Here is the Python program you requested.

class Solution:
    def mctFromLeafValues(self, arr: List[int]) -> int:
        if not arr: return 0
        
        res = []

        while len(arr) > 1:
            temp_res = []
            temp_res = [arr[i]*arr[i+1] for i in range(len(arr)-1)]
            idx = temp_res.index(min(temp_res))

            res.append(temp_res[idx])
            arr.pop(idx if arr[idx] < arr[idx+1] else idx+1)
                
            # left = arr[0] * arr[1]
            # right = arr[-1] * arr[-2]
            # if left < right:
            #     res.append(left)
            #     arr.pop(1 if arr[1] < arr[0] else 0)
            # elif right < left:
            #     res.append(right)
            #     arr.pop(-2 if arr[-2] < arr[-1] else -1)
            # else:
            #     res.append(left)
            #     if max(arr[0], arr[1]) > max(arr[-1], arr[-2]):
            #         arr.pop(-2 if arr[-2] < arr[-1] else -1)
            #     else:
            #         arr.pop(1 if arr[1] < arr[0] else 0)
        
        return sum(res)