<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>

int dp[41][41];
class Solution {
public:
    int asb(vector<int> &arr, int i, int j) {
        if(i==j) {
            return arr[i];
        } 
        if(dp[i][j]!=-1) {
            return dp[i][j];
        }
        else {
            int ans = INT_MAX;
            vector<int> d(j-i+1);
            int mx = INT_MIN;
            for(int k=j-i;k>=0;k--) {
                mx = max(mx, arr[k+i]);
                d[k] = mx;
            }
            mx = arr[i];
            for(int k=i;k<j;k++) {
                mx = max(arr[k], mx);
               ans = min(ans, mx*d[k+1-i]+asb(arr, i, k) + asb(arr, k+1, j));
            }
            return dp[i][j] =  ans;
        }
    }
    int mctFromLeafValues(vector<int>& arr) {
        int sum = 0;
        memset(dp, -1, sizeof(dp));
        for(auto i: arr) {
            sum+=i;
        }

        return asb(arr, 0, arr.size()-1) - sum;
    }
};
13600