<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 res = 0;
    while (A.size() > 1) {
        // Find the index of the minimum element
        auto it = min_element(A.begin(), A.end());
        int i = distance(A.begin(), it);

        // Find the smaller of the neighbors (A[i-1] and A[i+1] if they exist)
        int left = (i > 0) ? A[i - 1] : INT_MAX;
        int right = (i < A.size() - 1) ? A[i + 1] : INT_MAX;
        res += min(left, right) * A[i];

        // Remove the element at index i
        A.erase(A.begin() + i);
    }
    return res;  
    }
};
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