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Minimum Cost Tree From Leaf Values

Medium

MathWorks

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Topics:

ArraysDynamic Programming

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.
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.

A node is a leaf if and only if it has zero children.

Example 1:

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

Example 2:

Input: arr = [4,11]
Output: 44

Constraints:

2 <= arr.length <= 40
1 <= arr[i] <= 15
It is guaranteed that the answer fits into a 32-bit signed integer (i.e., it is less than 2³¹).

Try coding it in LeetCode

Minimum Cost Tree From Leaf Values

Medium

MathWorks

4 views

Topics:

ArraysDynamic Programming

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.
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.

A node is a leaf if and only if it has zero children.

Example 1:

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

Example 2:

Input: arr = [4,11]
Output: 44

Constraints:

2 <= arr.length <= 40
1 <= arr[i] <= 15
It is guaranteed that the answer fits into a 32-bit signed integer (i.e., it is less than 2³¹).

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array `arr`? Specifically, what is the upper bound on the number of leaf nodes?
Can the input array `arr` contain duplicate values, and if so, how should that affect the construction of the minimum cost tree?
Is the input array `arr` guaranteed to contain only positive integers, or could it contain zero or negative values?
If the input array `arr` is empty or contains only one element, what should the function return?
Could you provide an example of a non-trivial input and the expected output to further clarify the problem's constraints?

Brute Force Solution

Approach

The brute force approach to finding the minimum cost tree from leaf values is like trying out every single possible tree you can make with those leaf values. We calculate the cost for each possible tree and then pick the one with the lowest cost.

Here's how the algorithm would work step-by-step:

First, consider all possible ways to split the list of leaf values into two smaller lists.
For each split, calculate the cost of merging those two lists into a single node in the tree. This cost is the product of the largest value in each of the two lists.
Then, for each of the smaller lists we created, repeat the process: try every possible split of that list into two even smaller lists and calculate the cost of merging those.
Keep splitting and merging lists until you have individual leaf values. Each merge represents a node in a potential tree.
For each tree you've built this way, add up all the merging costs you calculated along the way to get the total cost for that tree.
Finally, compare the total costs of all the possible trees you created and choose the tree with the smallest total cost. That's your minimum cost tree.

Code Implementation

def minimum_cost_tree_from_leaf_values_brute_force(leaf_values):

    def calculate_minimum_cost(current_leaf_values):
        number_of_leafs = len(current_leaf_values)
        if number_of_leafs <= 1:
            return 0

        minimum_cost = float('inf')

        # Try all possible splits of the leaf values.
        for i in range(1, number_of_leafs):
            left_subtree = current_leaf_values[:i]
            right_subtree = current_leaf_values[i:]

            # The cost of merging the subtrees is the product of their maximum values.
            merging_cost = max(left_subtree) * max(right_subtree)

            # Recursively calculate the cost of the subtrees and add them.
            left_subtree_cost = calculate_minimum_cost(left_subtree)

            right_subtree_cost = calculate_minimum_cost(right_subtree)

            total_cost = merging_cost + left_subtree_cost + right_subtree_cost
            minimum_cost = min(minimum_cost, total_cost)

        return minimum_cost

    return calculate_minimum_cost(leaf_values)

Big(O) Analysis

Time Complexity

O(n^3) – The brute force approach explores all possible binary trees formed from the leaf values. For an array of size n, we consider all possible splits, resulting in approximately 2^(n-1) potential binary trees. Each split involves finding the maximum value in the left and right sub-arrays, which takes O(n) time. Since we explore a tree-like structure of splits, the depth of this tree is O(n) in the worst case. Therefore, the total number of operations becomes proportional to the number of possible trees multiplied by the cost of splitting each tree level, leading to O(n * 2^n) complexity. However, the described approach effectively iterates through all sub-arrays (of which there are O(n^2)), and for each sub-array, finds the minimum cost to split it into two, leading to O(n) operations for finding minimum cost. Therefore, multiplying the sub-array count and min cost operations, the complexity will be O(n^3).

Space Complexity

O(N^2) – The brute force approach, as described, explores all possible binary trees. This involves creating numerous sublists and storing the intermediate costs calculated for each split. Because every sublist split is considered, the number of calls grows to a point where all combinations of i and j indexes between 0 and N are being calculated. This leads to storing costs in a memoization array of size N x N for each subproblem to prevent redundant calculation, which dominates the space complexity, resulting in O(N^2) auxiliary space, where N is the number of leaf values.

Optimal Solution

Approach

The key is to build the tree from the bottom up. We repeatedly combine the smallest pairs of leaves, calculating the cost of each merge, until we're left with a single root.

Here's how the algorithm would work step-by-step:

Think of the problem as repeatedly merging pairs of numbers until only one remains.
At each step, find the smallest adjacent pair of numbers.
Multiply these two numbers to get the cost of merging them.
Replace the pair with the larger of the two numbers. This is important: we only keep the larger one because it affects future calculations.
Add the cost to a running total.
Repeat steps 2-5 until only one number remains.
The running total represents the minimum cost to build the tree.

Code Implementation

def minimum_cost_tree_from_leaf_values(leaf_values): 
    minimum_cost = 0
    while len(leaf_values) > 1:
        minimum_index = 0
        
        #Find the index of the smallest adjacent pair
        for index in range(1, len(leaf_values)): 
            if leaf_values[index] * leaf_values[index-1] < \
               leaf_values[minimum_index] * leaf_values[minimum_index-1]:
                minimum_index = index
        
        #Cost is product of the smallest adjacent pair
        minimum_cost += leaf_values[minimum_index] * leaf_values[minimum_index - 1]
        
        #Replace pair with larger element
        leaf_values[minimum_index-1] = max(leaf_values[minimum_index], leaf_values[minimum_index-1])

        #Remove the smaller element
        leaf_values.pop(minimum_index)

    return minimum_cost

Big(O) Analysis

Time Complexity

O(n³) – The algorithm involves repeatedly finding the minimum adjacent pair in an array of size n. Finding this pair requires scanning the array, which takes O(n) time. After finding the pair, we merge them by removing one element and updating another, which also takes O(n) time for shifting elements in the worst case. Since we repeat this merging process until only one element remains, we perform this O(n) operation n-1 times. Therefore the total time complexity is approximately n * (n+n) which simplifies to O(n³).

Space Complexity

O(N) – The described algorithm repeatedly modifies the input list and may use a temporary space to store the modified list during the merge process. In each merge operation, the pair of numbers is replaced by the larger of the two, potentially creating a new list or modifying the existing one in place, where N is the initial size of the input array. The space complexity is thus proportional to the number of elements in the input array, leading to O(N) space complexity because we are potentially holding a modified copy or performing in-place manipulation which in the worst case would require copying elements.

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no leaf nodes and thus no non-leaf nodes to sum.
Input array with only one element	Return 0 as a single leaf node forms a trivial tree with no non-leaf nodes.
Input array with two elements	Return the product of the two elements, which is the only possible cost.
Input array with all elements being the same value	The algorithm should still compute the correct minimum cost, as the leaf values are handled uniformly.
Input array with elements sorted in increasing order	The algorithm should efficiently explore different tree structures despite the sorted input.
Input array with elements sorted in decreasing order	The algorithm should efficiently explore different tree structures despite the reverse-sorted input.
Input array with very large integer values leading to potential integer overflow	Use a data type that can accommodate larger products, such as long or consider using modular arithmetic if appropriate and stated by problem.
Input array with a large number of elements	Dynamic programming optimization is required to prevent exceeding time limit due to exponential tree growth.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array `arr`? Specifically, what is the upper bound on the number of leaf nodes?
Can the input array `arr` contain duplicate values, and if so, how should that affect the construction of the minimum cost tree?
Is the input array `arr` guaranteed to contain only positive integers, or could it contain zero or negative values?
If the input array `arr` is empty or contains only one element, what should the function return?
Could you provide an example of a non-trivial input and the expected output to further clarify the problem's constraints?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

First, consider all possible ways to split the list of leaf values into two smaller lists.
For each split, calculate the cost of merging those two lists into a single node in the tree. This cost is the product of the largest value in each of the two lists.
Then, for each of the smaller lists we created, repeat the process: try every possible split of that list into two even smaller lists and calculate the cost of merging those.
Keep splitting and merging lists until you have individual leaf values. Each merge represents a node in a potential tree.
For each tree you've built this way, add up all the merging costs you calculated along the way to get the total cost for that tree.
Finally, compare the total costs of all the possible trees you created and choose the tree with the smallest total cost. That's your minimum cost tree.

Code Implementation

def minimum_cost_tree_from_leaf_values_brute_force(leaf_values):

    def calculate_minimum_cost(current_leaf_values):
        number_of_leafs = len(current_leaf_values)
        if number_of_leafs <= 1:
            return 0

        minimum_cost = float('inf')

        # Try all possible splits of the leaf values.
        for i in range(1, number_of_leafs):
            left_subtree = current_leaf_values[:i]
            right_subtree = current_leaf_values[i:]

            # The cost of merging the subtrees is the product of their maximum values.
            merging_cost = max(left_subtree) * max(right_subtree)

            # Recursively calculate the cost of the subtrees and add them.
            left_subtree_cost = calculate_minimum_cost(left_subtree)

            right_subtree_cost = calculate_minimum_cost(right_subtree)

            total_cost = merging_cost + left_subtree_cost + right_subtree_cost
            minimum_cost = min(minimum_cost, total_cost)

        return minimum_cost

    return calculate_minimum_cost(leaf_values)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

The key is to build the tree from the bottom up. We repeatedly combine the smallest pairs of leaves, calculating the cost of each merge, until we're left with a single root.

Here's how the algorithm would work step-by-step:

Think of the problem as repeatedly merging pairs of numbers until only one remains.
At each step, find the smallest adjacent pair of numbers.
Multiply these two numbers to get the cost of merging them.
Replace the pair with the larger of the two numbers. This is important: we only keep the larger one because it affects future calculations.
Add the cost to a running total.
Repeat steps 2-5 until only one number remains.
The running total represents the minimum cost to build the tree.

Code Implementation

def minimum_cost_tree_from_leaf_values(leaf_values): 
    minimum_cost = 0
    while len(leaf_values) > 1:
        minimum_index = 0
        
        #Find the index of the smallest adjacent pair
        for index in range(1, len(leaf_values)): 
            if leaf_values[index] * leaf_values[index-1] < \
               leaf_values[minimum_index] * leaf_values[minimum_index-1]:
                minimum_index = index
        
        #Cost is product of the smallest adjacent pair
        minimum_cost += leaf_values[minimum_index] * leaf_values[minimum_index - 1]
        
        #Replace pair with larger element
        leaf_values[minimum_index-1] = max(leaf_values[minimum_index], leaf_values[minimum_index-1])

        #Remove the smaller element
        leaf_values.pop(minimum_index)

    return minimum_cost

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 as there are no leaf nodes and thus no non-leaf nodes to sum.
Input array with only one element	Return 0 as a single leaf node forms a trivial tree with no non-leaf nodes.
Input array with two elements	Return the product of the two elements, which is the only possible cost.
Input array with all elements being the same value	The algorithm should still compute the correct minimum cost, as the leaf values are handled uniformly.
Input array with elements sorted in increasing order	The algorithm should efficiently explore different tree structures despite the sorted input.
Input array with elements sorted in decreasing order	The algorithm should efficiently explore different tree structures despite the reverse-sorted input.
Input array with very large integer values leading to potential integer overflow	Use a data type that can accommodate larger products, such as long or consider using modular arithmetic if appropriate and stated by problem.
Input array with a large number of elements	Dynamic programming optimization is required to prevent exceeding time limit due to exponential tree growth.