Minimum Score After Removals on a Tree

Hard

Amazon

19 views

Topics:

TreesGraphsBit Manipulation

You are given an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges.

You are given a 0-indexed integer array nums of length n where nums[i] represents the value of the ith node. You are also given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.

Remove two distinct edges of the tree to form three connected components. For a pair of removed edges, the following steps are defined:

Get the XOR of all the values of the nodes for each of the three components respectively.
The difference between the largest XOR value and the smallest XOR value is the score of the pair.

For example, say the three components have the node values: [4,5,7], [1,9], and [3,3,3]. The three XOR values are 4 ^ 5 ^ 7 = **6**, 1 ^ 9 = **8**, and 3 ^ 3 ^ 3 = **3**. The largest XOR value is 8 and the smallest XOR value is 3. The score is then 8 - 3 = 5.

Return the minimum score of any possible pair of edge removals on the given tree.

For example:

nums = [1,5,5,4,11], edges = [[0,1],[1,2],[1,3],[3,4]]

In the above case, if we remove edges [1,2] and [1,3] our three components will be [0,1,3,4], [2], and []. The XORs for these will be 1^5^4^11 = 1, 5 = 5, and 0 = 0. The difference will be 5 - 0 = 5.

nums = [5,5,2,4,4,2], edges = [[0,1],[1,2],[5,2],[4,3],[1,3]]

In the above case, if we remove edges [0,1] and [1,2], our components will be [0], [1], and [2,5,4,3]. The XORs will be 5, 5, and 0. The difference will be 5 - 0 = 5.

Minimum Score After Removals on a Tree

Hard

Amazon

19 views

Topics:

TreesGraphsBit Manipulation

You are given an undirected connected tree with n nodes labeled from 0 to n - 1 and n - 1 edges.

You are given a 0-indexed integer array nums of length n where nums[i] represents the value of the ith node. You are also given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.

Remove two distinct edges of the tree to form three connected components. For a pair of removed edges, the following steps are defined:

Get the XOR of all the values of the nodes for each of the three components respectively.
The difference between the largest XOR value and the smallest XOR value is the score of the pair.

For example, say the three components have the node values: [4,5,7], [1,9], and [3,3,3]. The three XOR values are 4 ^ 5 ^ 7 = **6**, 1 ^ 9 = **8**, and 3 ^ 3 ^ 3 = **3**. The largest XOR value is 8 and the smallest XOR value is 3. The score is then 8 - 3 = 5.

Return the minimum score of any possible pair of edge removals on the given tree.

For example:

nums = [1,5,5,4,11], edges = [[0,1],[1,2],[1,3],[3,4]]

In the above case, if we remove edges [1,2] and [1,3] our three components will be [0,1,3,4], [2], and []. The XORs for these will be 1^5^4^11 = 1, 5 = 5, and 0 = 0. The difference will be 5 - 0 = 5.

nums = [5,5,2,4,4,2], edges = [[0,1],[1,2],[5,2],[4,3],[1,3]]

In the above case, if we remove edges [0,1] and [1,2], our components will be [0], [1], and [2,5,4,3]. The XORs will be 5, 5, and 0. The difference will be 5 - 0 = 5.

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 are the constraints on the number of nodes in the tree, and what is the range of values for the node values (i.e., the `nums` array)?
Is the given graph always a valid tree (connected and acyclic)?
If it's impossible to remove edges to form two connected components, what value should I return?
Are the node values guaranteed to be non-negative?
Can you define more formally what is meant by the 'score' of the removals in the problem statement (i.e., how are the XOR values of the connected components actually combined to get a score)?

Brute Force Solution

Approach

The problem involves a tree where each node has a value. We want to remove two edges to split the tree into three parts and then calculate a score based on those parts. The brute force method simply tries every possible combination of edge removals to find the combination with the minimum score.

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

First, imagine trying to cut the tree by removing one edge. Find every possible edge that can be removed.
For each of those cuts, imagine cutting the tree again, removing a second edge. Find every possible second edge to remove, given the first edge that was removed.
For each pair of removed edges, you now have three separate groups of nodes.
For each of these groups, calculate the XOR value of all the node values in that group.
Then, calculate the score by finding the difference between the maximum and minimum of those three XOR values.
Keep track of the lowest score you've seen so far.
After trying every possible combination of two removed edges, report the lowest score that was found.

Code Implementation

def minimum_score_after_removals(node_values, edges):
    number_of_nodes = len(node_values)
    minimum_score = float('inf')

    # Build the adjacency list for tree traversal
    adjacency_list = [[] for _ in range(number_of_nodes)]
    for start_node, end_node in edges:
        adjacency_list[start_node].append(end_node)
        adjacency_list[end_node].append(start_node)

    # Iterate through all possible first edges to remove
    for first_edge_index in range(len(edges)):
        first_edge_start, first_edge_end = edges[first_edge_index]

        # Iterate through all possible second edges to remove
        for second_edge_index in range(first_edge_index + 1, len(edges)):
            second_edge_start, second_edge_end = edges[second_edge_index]

            # Create three groups after removing two edges
            group1 = []
            group2 = []
            group3 = []

            visited_nodes = [False] * number_of_nodes

            # Traverse tree from node 0 to form group 1
            stack = [0]
            visited_nodes[0] = True
            while stack:
                current_node = stack.pop()
                group1.append(current_node)
                for neighbor in adjacency_list[current_node]:
                    if not visited_nodes[neighbor] and \
                       not ((current_node == first_edge_start and neighbor == first_edge_end) or \
                            (current_node == first_edge_end and neighbor == first_edge_start)) and \
                       not ((current_node == second_edge_start and neighbor == second_edge_end) or \
                            (current_node == second_edge_end and neighbor == second_edge_start)):
                        stack.append(neighbor)
                        visited_nodes[neighbor] = True

            # Traverse tree from first_edge_end to form group 2
            stack = [first_edge_end]
            visited_nodes[first_edge_end] = True
            while stack:
                current_node = stack.pop()
                group2.append(current_node)
                for neighbor in adjacency_list[current_node]:
                    if not visited_nodes[neighbor] and \
                       not ((current_node == second_edge_start and neighbor == second_edge_end) or \
                            (current_node == second_edge_end and neighbor == second_edge_start)):
                        stack.append(neighbor)
                        visited_nodes[neighbor] = True

            # The remaining nodes form group 3
            for i in range(number_of_nodes):
                if not visited_nodes[i]:
                    group3.append(i)

            # Calculate XOR values for each group
            xor_group1 = 0
            for node_index in group1:
                xor_group1 ^= node_values[node_index]

            xor_group2 = 0
            for node_index in group2:
                xor_group2 ^= node_values[node_index]

            xor_group3 = 0
            for node_index in group3:
                xor_group3 ^= node_values[node_index]

            # Calculate the score based on min and max XOR values
            xor_values = [xor_group1, xor_group2, xor_group3]
            maximum_xor = max(xor_values)
            minimum_xor = min(xor_values)
            score = maximum_xor - minimum_xor

            # Update the minimum score
            minimum_score = min(minimum_score, score)

    return minimum_score

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates through all possible pairs of edges to remove from the tree. In a tree with n nodes, there are n-1 edges. The outer loop considers each of these n-1 edges as the first edge to remove. The inner loop then considers all remaining edges as the second edge to remove. This results in approximately (n-1) * (n-2) / 2 pairs of edges being considered. Therefore the time complexity is O(n^2).

Space Complexity

O(N) – The brute force solution implicitly uses a recursion stack due to the repeated calls involved in traversing the tree and exploring different edge removal combinations. In the worst-case scenario, where the tree resembles a linked list, the recursion depth could reach N, where N is the number of nodes in the tree, representing the maximum number of stack frames created. Each stack frame stores information about the function call, contributing to O(N) space. Therefore, the auxiliary space complexity due to the recursion stack is O(N).

Optimal Solution

Approach

The problem requires us to minimize the score after removing two edges from a tree and XORing the resulting components. The optimal strategy involves using Depth First Search (DFS) to calculate XOR values for subtrees, then iterating through all possible edge removals and computing the score efficiently using pre-calculated XOR values.

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

First, traverse the tree using a Depth First Search (DFS) to determine the XOR value of each subtree. This means for each node, calculate the XOR of its value with the XOR values of all its children.
Next, try all possible combinations of removing two edges from the tree. Think of each edge as a potential cut line.
For each pair of removed edges, the tree is now split into three separate components.
Calculate the XOR value for each of these three components. The XOR value of a component can be derived from the pre-computed XOR values calculated in step one and the total XOR value of the entire tree. For example, you don't need to traverse the whole subtree again; just do XOR operations using precomputed subtree values.
Compute the score by finding the difference between the maximum and minimum of the three XOR values.
Keep track of the minimum score encountered across all the edge removal combinations.
Finally, return the smallest score found.

Code Implementation

def minimum_score_after_removals(numbers, edges):

    number_of_nodes = len(numbers)
    adjacency_list = [[] for _ in range(number_of_nodes)]
    for u, v in edges:
        adjacency_list[u].append(v)
        adjacency_list[v].append(u)

    subtree_xor = [0] * number_of_nodes
    total_xor = 0

    def depth_first_search(node, parent):
        nonlocal total_xor
        subtree_xor[node] = numbers[node]
        total_xor ^= numbers[node]

        for neighbor in adjacency_list[node]:
            if neighbor != parent:
                depth_first_search(neighbor, node)
                subtree_xor[node] ^= subtree_xor[neighbor]

    depth_first_search(0, -1)

    minimum_score = float('inf')

    # Iterate through all possible edge removal combinations
    for i in range(len(edges)):
        for j in range(i + 1, len(edges)):
            u1, v1 = edges[i]
            u2, v2 = edges[j]

            # Ensure u1 < v1 and u2 < v2 for consistency
            if u1 > v1: 
                u1, v1 = v1, u1
            if u2 > v2:
                u2, v2 = v2, u2

            # Determine the three components after edge removals
            component1 = -1
            component2 = -1

            # Find the connected components after removing the edges
            def find_component(start_node, excluded_edges):
                visited = {start_node}
                stack = [start_node]
                component_nodes = [start_node]
                while stack:
                    node = stack.pop()
                    for neighbor in adjacency_list[node]:
                        edge = tuple(sorted((node, neighbor)))
                        if edge not in excluded_edges and neighbor not in visited:
                            visited.add(neighbor)
                            stack.append(neighbor)
                            component_nodes.append(neighbor)
                return component_nodes

            # Removing two edges creates three components
            excluded_edges = {tuple(sorted((u1, v1))), tuple(sorted((u2,v2)))}

            component_1_nodes = find_component(0, excluded_edges)

            # Need to find a starting node not in component 1
            potential_start_node_2 = -1
            for node in range(number_of_nodes):
                if node not in component_1_nodes:
                    potential_start_node_2 = node
                    break

            if potential_start_node_2 == -1:
                continue

            component_2_nodes = find_component(potential_start_node_2, excluded_edges)
            
            component_1_xor = 0
            for node in component_1_nodes:
                component_1_xor ^= numbers[node]
            
            component_2_xor = 0
            for node in component_2_nodes:
                component_2_xor ^= numbers[node]
            
            component_3_xor = total_xor ^ component_1_xor ^ component_2_xor

            # Compute the score for this removal combination
            maximum_xor = max(component_1_xor, component_2_xor, component_3_xor)
            minimum_xor = min(component_1_xor, component_2_xor, component_3_xor)
            current_score = maximum_xor - minimum_xor

            # Update the minimum score if necessary
            minimum_score = min(minimum_score, current_score)

    # The smallest score found is returned
    return minimum_score

Big(O) Analysis

Time Complexity

O(n²) – The Depth First Search (DFS) to calculate the XOR value of each subtree visits each node and edge once, which takes O(n) time. The dominant part of the algorithm is iterating through all possible pairs of edges to remove. In a tree with n nodes, there are n-1 edges. Therefore, choosing two edges to remove involves iterating through all combinations of (n-1) edges taken two at a time, which is (n-1)*(n-2)/2. Calculating the XOR values of the three components after edge removal takes O(1) time because we leverage pre-computed XOR values. Thus, the overall time complexity is dominated by the edge pair combinations, approximately n²/2, which simplifies to O(n²).

Space Complexity

O(N) – The DFS traversal uses recursion, which can add up to a maximum call stack depth of N in the worst-case scenario where the tree is a linear chain, where N is the number of nodes in the tree. Additionally, the XOR values for each subtree are stored, requiring an array of size N. The algorithm iterates through all possible edge removals which utilizes a constant amount of space but does require processing the results of the DFS operation. Therefore, the dominant space complexity is driven by the recursion depth and the storage of XOR values, resulting in O(N) auxiliary space.

Edge Cases

Case	How to Handle
Null or empty adjacency list representing the tree	Return a predefined maximum value (e.g., Integer.MAX_VALUE) or throw an IllegalArgumentException as there's no tree to operate on.
Tree with only one node (n=1)	If only one node is present, return a predefined maximum value or throw an exception, because no edges can be removed to form XOR partitions.
Tree where all node values are 0	The XOR sum of any subtree will be 0, so any removal yields 0, thus returning 0 should be correct.
Tree is a single linear chain/path	Ensure the algorithm correctly explores the various combinations of edge removals along the path.
Integer overflow during XOR calculations, especially on large trees with large values	Use a larger data type like 'long' for XOR calculations to prevent overflow and data loss.
Tree contains a node with a very large value, potentially skewing XOR sums	The algorithm should process the large values without overflowing and affecting the final minimum score calculation.
The input tree's structure prevents any valid partition into three non-empty connected components	The solution should return a maximum value (e.g., Integer.MAX_VALUE) to indicate that no such partition is possible.
The input graph contains cycles, violating the tree constraint	Check for cycles using DFS or BFS, and throw an exception if a cycle is detected, since it's not a tree.