1130. 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

Constraints:

• 2 <= arr.length <= 40

• 1 <= arr[i] <= 15

• It is guaranteed that the answer fits into a 32-bit signed integer (ie. it is less than 2^31).

// Monotonic Decreasing Stack
int mctFromLeafValues(vector<int>& arr) { // time: O(n); space: O(n)
    int res = 0, n = arr.size();
    vector<int> stack({INT_MAX});
    for (int a : arr) {
        while (stack.back() <= a) {
            int mid = stack.back();
            stack.pop_back();
            res += mid * min(a, stack.back());
        }
        stack.push_back(a);
    }
    for (int i = 2; i < stack.size(); ++i) {
        res += stack[i - 1] * stack[i];
    }
    return res;
}

// Top-Down Dynamic Programming
// dp(left, right)= min( max(arr[left ... i]) * max(arr[i+1 ... right]) + dp(left, i) + dp(i + 1, right) ) 
// where i go from left to right-1
int dp(int left, int right, vector<vector<int> >& memo, vector<vector<int> >& maxi) {
    if (left == right) return 0; // leaf
    if (memo[left][right] != -1) return memo[left][right];
    int res = INT_MAX;
    for (int i = left; i < right; ++i) {
        res = min(res, maxi[left][i] * maxi[i + 1][right] + dp(left, i, memo, maxi) + dp(i + 1, right, memo, maxi));
    }
    return memo[left][right] = res;
}
int mctFromLeafValues(vector<int>& arr) { // time: O(n^3); space: O(n^2)
    int n = arr.size();
    // maxi[i][j]: records the maximal number in arr[i...j]
    vector<vector<int> > memo(n, vector<int>(n, -1)), maxi(n, vector<int>(n, 0));
    for(int i = 0; i < n; ++i){
        maxi[i][i] = arr[i];
        for (int j = i + 1; j < n; ++j) {
            maxi[i][j] = max(maxi[i][j - 1], arr[j]);
        }
    }
    return dp(0, arr.size() - 1, memo, maxi);
}

// Bottom-Up Dynamic Programming
int mctFromLeafValues(vector<int>& arr) { // time: O(n^3); space: O(n^2)
    int n = arr.size();
    vector<vector<int> > dp(n, vector<int>(n)), maxi(n, vector<int>(n));
    for (int i = 0; i < n; ++i) {
        int local_max = 0;
        for (int j = i; j < n; ++j) {
            if (arr[j] > local_max) {
                local_max = arr[j];
            }
            maxi[i][j] = local_max;
        }
    }
    for (int len = 1; len < n; ++len) {
        for (int left = 0; left + len < n; ++left) {
            int right = left + len;
            dp[left][right] = INT_MAX;
            if (len == 1) {
                dp[left][right] = arr[left] * arr[right];
            } else {
                for (int k = left; k < right; ++k) {
                    dp[left][right] = min(dp[left][right], 
                                          dp[left][k] + dp[k + 1][right] + maxi[left][k] * maxi[k + 1][right]);
                }
            }
        }
    }
    return dp[0][n - 1];
}