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

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.

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)