<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 mctFromLeafValues(vector<int>& A) {
    //     int n=a.size();
    //     vector<vector<int>>dp(n,vector<int>(n,1e9));
    //     for(int i=0;i<n;i++)dp[i][i]=0;
    //     for(int d=1;d<n;d++){
    //     for(int i=0;i<n;i++){
    //         int j=i+d; if(j>=n)continue;
    //             for(int k=i;k<j;k++){
    //                 int mx1=0;
    //                 for(int p=i;p<=k;p++)mx1=max(mx1,a[p]);
    //                 int mx2=0;
    //                 for(int p=k+1;p<=j;p++)mx2=max(mx2,a[p]);
    //                 dp[i][j]=min(dp[i][j],mx1*mx2+dp[i][k]+dp[k+1][j]);
    //                 cout<<dp[i][j]<<" "<<i<<" "<<j<<"\n";
    //             }
            
    //     }
    // }
    //     return dp[0][n-1];
   int res = 0;
        vector<int> stack = {INT_MAX};
        for (int a : A) {
            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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