> For the complete documentation index, see [llms.txt](https://soumyajit4419.gitbook.io/ds-algo/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://soumyajit4419.gitbook.io/ds-algo/greedy-algorithms/13.-minimum-cost-tree-from-leaf-values.md).

# 13. Minimum Cost Tree From Leaf Values

Given an array `arr` of positive integers, consider all binary trees such that:

* Each node has either 0 or 2 children;
* The values of `arr` correspond to the values of each **leaf** in an in-order traversal of the tree.  *(Recall that a node is a leaf if and only if it has 0 children.)*
* The value of each non-leaf node is equal to the product of the largest leaf value in its left and right subtree respectively.

Among all possible binary trees considered, return the smallest possible sum of the values of each non-leaf node.  It is guaranteed this sum fits into a 32-bit integer.

**Example 1:**

```
Input: arr = [6,2,4]
Output: 32
Explanation:
There are two possible trees.  The first has non-leaf node sum 36, and the second has non-leaf node sum 32.

    24            24
   /  \          /  \
  12   4        6    8
 /  \               / \
6    2             2   4
```

## Solution : (Dp + Memo)

**Trying to split at each possible point**

```cpp
class Solution
{
public:
    int findVal(vector<int> &arr, int start, int end, vector<vector<int>> &dp)
    {

        if (start >= end)
        {
            return 0;
        }

        if (dp[start][end] != -1)
        {
            return dp[start][end];
        }

        int ans = INT_MAX;

        for (int i = start; i < end; i++)
        {
            int left = findVal(arr, start, i, dp);
            int right = findVal(arr, i + 1, end, dp);

            int leftMax = 0;
            int rightMax = 0;

            for (int j = start; j <= i; j++)
            {
                leftMax = max(leftMax, arr[j]);
            }

            for (int j = i + 1; j <= end; j++)
            {
                rightMax = max(rightMax, arr[j]);
            }

            ans = min(ans, left + right + leftMax * rightMax);
        }

        dp[start][end] = ans;
        return ans;
    }

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

        return findVal(arr, 0, n - 1, dp);
    }
};
```

## Solution: (Greedy)

**Approach:**\
**Finding the minimum product and the smaller element is just a special case of finding a valley.**\
**Finding the valley point i.e minimum value and multiplying it.**\
**Remove it from the array as it is the minimum.**

```cpp
class Solution
{
public:
    int mctFromLeafValues(vector<int> &arr)
    {

        int sum = 0;
        while (arr.size() > 1)
        {

            for (int i = 0; i < arr.size(); i++)
            {
                if (i == 0 && arr[i] <= arr[i + 1])
                {
                    sum += arr[i] * arr[i + 1];
                    arr.erase(arr.begin() + i);
                }
                else if (i == arr.size() - 1 && arr[i] <= arr[i - 1])
                {
                    sum += arr[i] * arr[i - 1];
                    arr.erase(arr.begin() + i);
                }
                else
                {
                    if (arr[i] <= arr[i + 1] && arr[i] <= arr[i - 1])
                    {
                        sum += (arr[i + 1] < arr[i - 1]) ? arr[i] * arr[i + 1] : arr[i] * arr[i - 1];
                        arr.erase(arr.begin() + i);
                    }
                }
            }
        }

        return sum;
    }
};
```

**Time Complexity: O(n^2)**


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