<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:
    map<pair<int,int>,pair<int,int>>dp;
    int mctFromLeafValues(vector<int>& arr) {
     int st = 0,en = arr.size()-1;
        
        auto ans = solve(st,en,arr);
        return ans.first;
    }
    pair<int,int> solve(int st, int en, vector<int>& arr)
    {
        // you can create different root nodes by doing n-1 partitions where n is the size of the arr.  for each partition of leaf nodes calculate the sum and max leaf 
        //compare the n-1 sum values for n-1 trees.  after the loop return the minsum 
        
        if(st == en)
            return {0,arr[st]};
        if(dp.count({st,en}))
            return dp[{st,en}];
        int minsum = INT_MAX,maxleaf = INT_MIN;
        for(int i = st; i < en; i++)
        {
            auto leftinfo = solve(st,i,arr);
            auto rightinfo = solve(i+1,en,arr);
            int sum  = leftinfo.first+ rightinfo.first + (leftinfo.second*rightinfo.second);
            minsum = min(minsum,sum);
            maxleaf = max(leftinfo.second,rightinfo.second);
        }
        return dp[{st,en}] = {minsum,maxleaf};
    }
};
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