<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 solve(vector<int>& arr,map<pair<int,int>,int> &maxi  , int left , int right,vector<vector<int>> &dp){
        //leaf node
        if(left==right){
            return 0;
        }

        if(dp[left][right]!=-1){
            return dp[left][right];
        }

        int ans = INT_MAX;

        for(int i = left ; i<right ;i++){
            ans = min(ans,maxi[{left,i}]*maxi[{i+1,right}] + solve(arr,maxi,left,i,dp)+ solve(arr,maxi,i+1,right,dp));
        }
        return  dp[left][right] = ans;
    }

    int solveTab(vector<int> &arr,map<pair<int,int>,int> &maxi){
        int n=arr.size();
        vector<vector<int>> dp(n+1,vector<int>(n+1,0));
        for(int left=n-1;left>=0;left--){
            for(int right=left;right<n;right++){
                if(left==right){
                    continue;
                }
                else{
                    int ans=INT_MAX;
                    for(int i=left;i<right;i++){
                        ans=min(ans,maxi[{left,i}]*maxi[{i+1,right}]+dp[left][i]+dp[i+1][right]);
                    }
                    dp[left][right]=ans;
                }
            }
        }
        return dp[0][n-1];
    }


    int mctFromLeafValues(vector<int>& arr) {
        // map<pair<int,int>,int> maxi;
        // int n = arr.size();
        // //vector<vector<int>> dp(n+1,vector<int>(n+1,-1));

        // for(int i = 0 ; i<arr.size();i++){
        //     maxi[{i,i}] = arr[i];
        //     for(int j = i+1 ; j<arr.size();j++){
        //         maxi[{i,j}] = max(arr[j], maxi[{i,j-1}]);
        //     }
        // }

        
    
        // //return solve(arr,maxi,0,arr.size()-1,dp);
        // return solveTab(arr,maxi);
        int res = 0;
        vector<int> stack = {INT_MAX};
        for (int a : arr) {
            while (stack.back() <= a) {
                int mid = stack.back();
                stack.pop_back();
                res += mid * min(stack.back(), a);
            }
            stack.push_back(a);
        }
        for (int i = 2; i < stack.size(); ++i) {
            res += stack[i] * stack[i - 1];
        }
        return res;
    }
};
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