<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 dp[105][105];

    int mctFromLeafValues(vector<int>& arr) {
        memset(dp,-1,sizeof(dp));
        map<int,int> mleft;
        map<int,int> mright;

        int n=arr.size();

        int leftmx=0;

        for(int i=0;i<n;i++){
            leftmx=max(leftmx,arr[i]);
            mleft[i]=leftmx;
        }


        int rightmx=0;

        for(int i=n-1;i>-1;i--){
            rightmx=max(rightmx,arr[i]);
            mright[i]=rightmx;
        }

        map<pair<int, int>, int> mp;
       
        for(int i=0;i<n;i++)
        {
            mp[{i,i}] = arr[i];
            for(int j=i+1;j<n;j++)
            {
                mp[{i,j}] = max(mp[{i,j-1}],arr[j]);
            }
        }

        return solve(arr,0,arr.size()-1,mleft,mright,mp);
    }

    int solve(vector<int> &arr,int i,int j,map<int,int> &mleft,map<int,int> &mright, map<pair<int, int>, int> &mp){
        if(i==j){
            return 0;
        }
        
        int ans=1e9;

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

        for(int k=i;k<j;k++){

            

            ans=min(ans,solve(arr,i,k,mleft,mright,mp)+solve(arr,k+1,j,mleft,mright,mp)+(mp[{i,k}]*mp[{k+1,j}]));
        }
        
        return dp[i][j]= ans;
    }
};
14500