<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 f(vector<int> &arr,int i,int j,vector<vector<int>> &dp){
        if(i>=j){
            return 0;
        }

        if(dp[i][j] != -1){
            return dp[i][j];
        }

        vector<int> maxi(j-i+1,0);
        int maxEle = INT_MIN;

        for(int k=j;k>=i;k--){
            maxEle = max(maxEle , arr[k]);

            maxi[k-i] = maxEle;
        }
        //now we know who are the maxes after a certain index

        int newMax = 0;
        int res = INT_MAX;

        for(int k=i;k<j;k++){
            newMax = max(newMax , arr[k]);
            // cout<<i<<" "<<k<<" "<<j<<" "<<maxi.size()<<endl;
            res = min(res , newMax*maxi[k+1-i] + f(arr,i,k,dp) + f(arr,k+1,j,dp) );
        }

        return dp[i][j] = res;
    }

    int mctFromLeafValues(vector<int>& arr) {
        int n = arr.size();
        vector<vector<int>> dp(n,vector<int>(n,-1));
        return f(arr,0,n-1,dp);
    }
};
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