<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:
//Let dp[i][j] represent the minimum sum of non-leaf node values for the subarray from index i to j.
//The cost for the subarray arr[i:j] will be the sum of the costs of the left and right subarrays plus the product of their maximum values.
//DP table to store the minimum sum of non-leaf nodes for the subarray from index i to j.
    int mctFromLeafValues(vector<int>& arr) {
         int n = arr.size();
        // Using long long to prevent overflow
        vector<vector<long long>> dp(n, vector<long long>(n, LLONG_MAX));
        vector<vector<int>> maxLeaf(n, vector<int>(n, 0));

        // Initialize maxLeaf
        for (int i = 0; i < n; ++i) {
            maxLeaf[i][i] = arr[i];
            dp[i][i] = 0; // Base case: No cost for single leaf
        }

        // Fill maxLeaf for all ranges
        for (int length = 2; length <= n; ++length) {
            for (int i = 0; i <= n - length; ++i) {
                int j = i + length - 1;
                maxLeaf[i][j] = max(maxLeaf[i][j-1], arr[j]);
            }
        }

        // Compute dp for all ranges
        for (int length = 2; length <= n; ++length) {
            for (int i = 0; i <= n - length; ++i) {
                int j = i + length - 1;
                for (int k = i; k < j; ++k) {
                    long long cost = dp[i][k] + dp[k+1][j] + (long long)maxLeaf[i][k] * maxLeaf[k+1][j];
                    dp[i][j] = min(dp[i][j], cost);
                }
            }
        }

        return dp[0][n-1];
    
        
    }
};
11600