<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>& arr) {
        int n = arr.size();
        vector<vector<int>> dp(n, vector<int>(n));

        for (int l = 1; l <= n; l++) {
            for (int i = 0; i <= n - l; i++) {
                int j = i + l - 1; 
                if (i == j) {
                    dp[i][j] = 0;
                    continue;
                }

                vector<int> maxVal(l);
                maxVal[l-1] = arr[j];

                for (int k = l - 2; k >= 0; k--) {
                    maxVal[k] = max(maxVal[k + 1], arr[i + k]);
                }

                int curMax = 0;

                dp[i][j] = INT_MAX;

                for (int k = i; k < j; k++) {
                    curMax = max(curMax, arr[k]);
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k+1][j] + curMax * maxVal[k + 1 - i]);
                }
            }
        }

        for (int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                cout << dp[i][j] << ' ';
            }
            cout << '\n';
        }

        return dp[0][n-1];
    }
};
12500