<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;
    pair<int, int> solve(int i, int j, vector<int> &arr) {
        // Base case
        if(i == j) return {0, arr[i]};
        if(j - i == 1) return {arr[i] * arr[j], max(arr[i], arr[j])};
        if(dp.count({i, j})) return dp[{i, j}];
        
        pair<int, int> ans = {INT_MAX, INT_MAX};
        for(int k = i; k < j; k++) {
            auto leftAns = solve(i, k, arr);
            auto rightAns = solve(k + 1, j, arr);
            int currAns = (leftAns.first + rightAns.first + (leftAns.second * rightAns.second));
            int currMax = max(leftAns.second, rightAns.second);
            if(currAns < ans.first) 
                ans = {currAns, currMax};
            else if(currAns == ans.first && currMax < ans.second)
                ans = {currAns, currMax};
        }
        return dp[{i, j}] = ans;
    }
    int mctFromLeafValues(vector<int>& arr) {
        return solve(0, arr.size() - 1, arr).first;
    }
};
14000