<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 dfs_max(vector<int> &nums, int left, int right, vector<vector<int>> &maxn) {
        if (left >= right) {
            maxn[left][right] = nums[left];
            return maxn[left][right];
        }
        if (maxn[left][right] != -1)
            return maxn[left][right];
        
        int res = 0;
        for (int i = left; i < right; i++) {
            res = max(res, max(dfs_max(nums, left, i, maxn), dfs_max(nums, i + 1, right, maxn)));
        }
        
        return maxn[left][right] = res;
    }
    
    int dfs(vector<int> &nums, int left, int right, vector<vector<int>> &dp, vector<vector<int>> &maxn) {
        if (left >= right)
            return 0;
        if (dp[left][right] != INT_MAX)
            return dp[left][right];
        
        int res = INT_MAX;
        for (int i = left; i < right; i++) {
            res = min(res, dfs(nums, left, i, dp, maxn) + dfs(nums, i + 1, right, dp, maxn) + maxn[left][i] * maxn[i + 1][right]);
        }
        
        return dp[left][right] = res;
    }
    
    int mctFromLeafValues(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> maxn(n, vector<int>(n, -1));
        vector<vector<int>> dp(n, vector<int>(n, INT_MAX));
        
        dfs_max(nums, 0, n - 1, maxn);
        
        return dfs(nums, 0, n - 1, dp, maxn);
    }
};
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