Minimum Cost Tree From Leaf Values
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1130.Minimum%20Cost%20Tree%20From%20Leaf%20Values/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1100-1199/1130.Minimum%20Cost%20Tree%20From%20Leaf%20Values/images/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>
Stack; Greedy; Array; Dynamic Programming; Monotonic Stack
class Solution {
    private Integer[][] f;
    private int[][] g;

    public int mctFromLeafValues(int[] arr) {
        int n = arr.length;
        f = new Integer[n][n];
        g = new int[n][n];
        for (int i = n - 1; i >= 0; --i) {
            g[i][i] = arr[i];
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = Math.max(g[i][j - 1], arr[j]);
            }
        }
        return dfs(0, n - 1);
    }

    private int dfs(int i, int j) {
        if (i == j) {
            return 0;
        }
        if (f[i][j] != null) {
            return f[i][j];
        }
        int ans = 1 << 30;
        for (int k = i; k < j; k++) {
            ans = Math.min(ans, dfs(i, k) + dfs(k + 1, j) + g[i][k] * g[k + 1][j]);
        }
        return f[i][j] = ans;
    }
}