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Maximum Product of Splitted Binary Tree

Medium

Amazon

2 views

Topics:

TreesRecursion

Given the root of a binary tree, split the binary tree into two subtrees by removing one edge such that the product of the sums of the subtrees is maximized.

Return the maximum product of the sums of the two subtrees. Since the answer may be too large, return it modulo 10^9 + 7.

Note that you need to maximize the answer before taking the mod and not after taking it.

For example:

Consider the binary tree represented by the array [1,2,3,4,5,6]. If we remove the edge between node 1 and node 2, we get two binary trees with sums 11 (3+6+1+4+5) and 10 (2). Their product is 110.

As another example, consider the binary tree [1,null,2,3,4,null,null,5,6]. If we remove the edge between node 2 and node 3, we get two binary trees with sums 15 and 6. Their product is 90.

Constraints:

The number of nodes in the tree is in the range [2, 5 * 10^4].
1 <= Node.val <= 10^4

Maximum Product of Splitted Binary Tree

Medium

Amazon

2 views

Topics:

TreesRecursion

Given the root of a binary tree, split the binary tree into two subtrees by removing one edge such that the product of the sums of the subtrees is maximized.

Return the maximum product of the sums of the two subtrees. Since the answer may be too large, return it modulo 10^9 + 7.

Note that you need to maximize the answer before taking the mod and not after taking it.

For example:

Consider the binary tree represented by the array [1,2,3,4,5,6]. If we remove the edge between node 1 and node 2, we get two binary trees with sums 11 (3+6+1+4+5) and 10 (2). Their product is 110.

As another example, consider the binary tree [1,null,2,3,4,null,null,5,6]. If we remove the edge between node 2 and node 3, we get two binary trees with sums 15 and 6. Their product is 90.

Constraints:

The number of nodes in the tree is in the range [2, 5 * 10^4].
1 <= Node.val <= 10^4

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 range of values for the nodes in the binary tree? Can they be negative or zero?
Is the input always a valid binary tree, or should I handle cases where the input might be malformed (e.g., a cycle, disconnected nodes)?
What should I return if the tree is empty (i.e., root is null)?
By 'splitted', do you mean removing a single edge, creating two separate subtrees? I want to confirm the interpretation of 'splitted'.
Is the goal to find the absolute maximum product, or is it possible for the product to overflow? If the product can be very large, should I return the result modulo some value?

Brute Force Solution

Approach

The brute force method for this tree problem involves trying every single way you can cut the tree into two parts. We'll calculate the product of the sums of these two parts for each possible cut, and then find the largest product we discovered.

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

First, find the total sum of all the numbers in the entire tree.
Next, go through each branch or connection in the tree, one at a time.
For each branch, imagine cutting the tree at that point. This will split the tree into two smaller trees.
Calculate the sum of the numbers in the smaller tree that is cut off.
The sum of the remaining part of the tree is simply the total sum of the entire tree, minus the sum of the smaller tree you just calculated.
Multiply these two sums together to get a product.
Keep track of the largest product you find as you try cutting at different branches.
After trying every single possible cut, the largest product you found is your answer.

Code Implementation

def maximum_product_splitted_binary_tree_brute_force(root):
    total_sum = calculate_tree_sum(root)
    maximum_product = 0

    def calculate_subtree_sum(node):
        if not node:
            return 0
        return node.val + calculate_subtree_sum(node.left) + calculate_subtree_sum(node.right)

    def traverse_tree(node):
        nonlocal maximum_product
        if not node:
            return

        if node.left:
            subtree_sum = calculate_subtree_sum(node.left)
            # Calculate product of the two split trees
            product = subtree_sum * (total_sum - subtree_sum)

            maximum_product = max(maximum_product, product)

        if node.right:
            subtree_sum = calculate_subtree_sum(node.right)
            # Calculate product of the two split trees
            product = subtree_sum * (total_sum - subtree_sum)

            maximum_product = max(maximum_product, product)

        traverse_tree(node.left)
        traverse_tree(node.right)

    def calculate_tree_sum(root):
        if not root:
            return 0
        return root.val + calculate_tree_sum(root.left) + calculate_tree_sum(root.right)

    traverse_tree(root)

    return maximum_product % (10**9 + 7)

Big(O) Analysis

Time Complexity

O(n) – The algorithm first calculates the total sum of the tree in O(n) time, where n is the number of nodes. Then, it iterates through each node (branch) of the tree, performing a split. For each split, it calculates the sum of the subtree, which can be done in O(n) in the worst case by traversing all nodes in the subtree, however, we only traverse the tree once and save the values using a postorder DFS. Since we are only traversing the tree once and calculating the values, the time complexity of calculating all subtree sums becomes O(n), so the overall time complexity is O(n).

Space Complexity

O(N) – The algorithm uses recursion to traverse the binary tree. In the worst-case scenario (e.g., a skewed tree), the recursion depth can be equal to the number of nodes in the tree, N. Each recursive call adds a new frame to the call stack. Therefore, the auxiliary space used by the call stack can grow linearly with the number of nodes, resulting in O(N) space complexity due to the recursion.

Optimal Solution

Approach

The goal is to cut the tree into two parts so that the product of their sums is as big as possible. The clever idea is to realize we don't need to explore every single possible cut; instead, we can first figure out the total sum of the whole tree, and then find a cut where one part is as close as possible to half of that total sum. This guarantees the other part will also be close to half, maximizing the product.

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

Calculate the total sum of all the values in the entire tree.
Now, think about cutting one edge of the tree. That cut will separate the tree into two pieces.
For each possible cut (edge) in the tree, calculate the sum of the smaller piece that gets cut off.
Find the cut where the sum of the smaller piece is closest to half of the total tree sum.
Once you've found that best cut, calculate the sum of the larger piece. You can do this by subtracting the sum of the smaller piece from the total tree sum.
Multiply the sum of the smaller piece by the sum of the larger piece. This is the largest product you can get by splitting the tree.
Return this maximum product.

Code Implementation

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def maxProduct(self, root: TreeNode) -> int:
        subtree_sums = []

        def calculate_subtree_sum(node):
            if not node:
                return 0
            
            left_sum = calculate_subtree_sum(node.left)
            right_sum = calculate_subtree_sum(node.right)
            
            current_subtree_sum = node.val + left_sum + right_sum
            subtree_sums.append(current_subtree_sum)
            return current_subtree_sum

        total_tree_sum = calculate_subtree_sum(root)

        max_product_so_far = 0

        # Iterate through subtree sums to find optimal split
        for subtree_sum in subtree_sums:
            max_product_so_far = max(max_product_so_far,
                                    subtree_sum * (total_tree_sum - subtree_sum))

        return max_product_so_far % (10**9 + 7)

Big(O) Analysis

Time Complexity

O(n) – The algorithm first calculates the total sum of the binary tree, which requires visiting each node once. Then, it traverses the tree again to find the sum of subtrees resulting from each possible cut. This also involves visiting each node once. Therefore, the time complexity is dominated by two tree traversals, both of which are linear with respect to the number of nodes n in the tree, resulting in O(n) time complexity.

Space Complexity

O(N) – The space complexity is dominated by the recursion stack used during the total sum calculation and the cut sum calculations. In the worst-case scenario, such as a skewed tree, the recursion depth can reach N, where N is the number of nodes in the tree. This leads to a stack frame for each node in the skewed tree. Therefore, the auxiliary space used by the recursion stack is proportional to N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null root node	Return 0 immediately as there's no tree to split.
Single node tree	Return 0 as you can't split a single node tree.
Tree with all nodes having value 0	Handle zero values correctly during sum and product calculations, avoiding division by zero issues and incorrect maximums.
Tree with large positive node values that cause integer overflow when multiplied	Use long data type to store intermediate products to prevent overflow.
Tree with negative node values	Negative values will affect the maximum product so include them in calculation and comparison.
Skewed tree (e.g., all nodes in the left subtree)	The solution should handle unbalanced trees without stack overflow in recursive calls by using efficient traversal methods.
Maximum size tree (deep and wide)	Ensure the algorithm has an acceptable time and space complexity (e.g., O(N) time and O(H) space, where N is the number of nodes and H is the height of the tree).
Integer.MAX_VALUE or Integer.MIN_VALUE node values	Ensure intermediate calculations involving these values do not cause overflow or unexpected behavior by using long for sums and products.

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 range of values for the nodes in the binary tree? Can they be negative or zero?
Is the input always a valid binary tree, or should I handle cases where the input might be malformed (e.g., a cycle, disconnected nodes)?
What should I return if the tree is empty (i.e., root is null)?
By 'splitted', do you mean removing a single edge, creating two separate subtrees? I want to confirm the interpretation of 'splitted'.
Is the goal to find the absolute maximum product, or is it possible for the product to overflow? If the product can be very large, should I return the result modulo some value?

Brute Force Solution

Approach

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

First, find the total sum of all the numbers in the entire tree.
Next, go through each branch or connection in the tree, one at a time.
For each branch, imagine cutting the tree at that point. This will split the tree into two smaller trees.
Calculate the sum of the numbers in the smaller tree that is cut off.
The sum of the remaining part of the tree is simply the total sum of the entire tree, minus the sum of the smaller tree you just calculated.
Multiply these two sums together to get a product.
Keep track of the largest product you find as you try cutting at different branches.
After trying every single possible cut, the largest product you found is your answer.

Code Implementation

def maximum_product_splitted_binary_tree_brute_force(root):
    total_sum = calculate_tree_sum(root)
    maximum_product = 0

    def calculate_subtree_sum(node):
        if not node:
            return 0
        return node.val + calculate_subtree_sum(node.left) + calculate_subtree_sum(node.right)

    def traverse_tree(node):
        nonlocal maximum_product
        if not node:
            return

        if node.left:
            subtree_sum = calculate_subtree_sum(node.left)
            # Calculate product of the two split trees
            product = subtree_sum * (total_sum - subtree_sum)

            maximum_product = max(maximum_product, product)

        if node.right:
            subtree_sum = calculate_subtree_sum(node.right)
            # Calculate product of the two split trees
            product = subtree_sum * (total_sum - subtree_sum)

            maximum_product = max(maximum_product, product)

        traverse_tree(node.left)
        traverse_tree(node.right)

    def calculate_tree_sum(root):
        if not root:
            return 0
        return root.val + calculate_tree_sum(root.left) + calculate_tree_sum(root.right)

    traverse_tree(root)

    return maximum_product % (10**9 + 7)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Calculate the total sum of all the values in the entire tree.
Now, think about cutting one edge of the tree. That cut will separate the tree into two pieces.
For each possible cut (edge) in the tree, calculate the sum of the smaller piece that gets cut off.
Find the cut where the sum of the smaller piece is closest to half of the total tree sum.
Once you've found that best cut, calculate the sum of the larger piece. You can do this by subtracting the sum of the smaller piece from the total tree sum.
Multiply the sum of the smaller piece by the sum of the larger piece. This is the largest product you can get by splitting the tree.
Return this maximum product.

Code Implementation

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def maxProduct(self, root: TreeNode) -> int:
        subtree_sums = []

        def calculate_subtree_sum(node):
            if not node:
                return 0
            
            left_sum = calculate_subtree_sum(node.left)
            right_sum = calculate_subtree_sum(node.right)
            
            current_subtree_sum = node.val + left_sum + right_sum
            subtree_sums.append(current_subtree_sum)
            return current_subtree_sum

        total_tree_sum = calculate_subtree_sum(root)

        max_product_so_far = 0

        # Iterate through subtree sums to find optimal split
        for subtree_sum in subtree_sums:
            max_product_so_far = max(max_product_so_far,
                                    subtree_sum * (total_tree_sum - subtree_sum))

        return max_product_so_far % (10**9 + 7)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null root node	Return 0 immediately as there's no tree to split.
Single node tree	Return 0 as you can't split a single node tree.
Tree with all nodes having value 0	Handle zero values correctly during sum and product calculations, avoiding division by zero issues and incorrect maximums.
Tree with large positive node values that cause integer overflow when multiplied	Use long data type to store intermediate products to prevent overflow.
Tree with negative node values	Negative values will affect the maximum product so include them in calculation and comparison.
Skewed tree (e.g., all nodes in the left subtree)	The solution should handle unbalanced trees without stack overflow in recursive calls by using efficient traversal methods.
Maximum size tree (deep and wide)	Ensure the algorithm has an acceptable time and space complexity (e.g., O(N) time and O(H) space, where N is the number of nodes and H is the height of the tree).
Integer.MAX_VALUE or Integer.MIN_VALUE node values	Ensure intermediate calculations involving these values do not cause overflow or unexpected behavior by using long for sums and products.