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Minimum Score After Removals on a Tree

Difficulty: Hard

Problem Description

There is 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:

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

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

Key Insights

  • The goal is to minimize the difference between the largest and smallest XOR values obtained from the three resulting components after removing two edges.
  • The structure of a tree allows for a unique path between any pair of nodes, which simplifies the problem of computing connected components.
  • Using Depth-First Search (DFS) can help in exploring the components effectively after edge removals.
  • The XOR operation has properties that can be leveraged to compute the required values efficiently.

Space and Time Complexity

Time Complexity: O(n^2) in the worst case, as we may need to check all pairs of edges and compute the XOR for the components. Space Complexity: O(n) for storing the adjacency list representation of the tree and the DFS stack.

Solution

To solve the problem, we will:

  1. Construct the tree using an adjacency list from the edges provided.
  2. For each pair of edges, remove them and use DFS to compute the XOR values of the resulting three components.
  3. Calculate the score for the pair of edges by finding the difference between the maximum and minimum XOR values.
  4. Keep track of the minimum score found across all pairs of edge removals.

Code Solutions

def minScore(nums, edges):
    from collections import defaultdict
    
    # Build the graph
    graph = defaultdict(list)
    for a, b in edges:
        graph[a].append(b)
        graph[b].append(a)

    def dfs(node, visited, values):
        visited.add(node)
        values[0] ^= nums[node]
        for neighbor in graph[node]:
            if neighbor not in visited:
                dfs(neighbor, visited, values)

    min_score = float('inf')
    n = len(nums)

    # Try removing each pair of edges
    for i in range(n - 1):
        for j in range(i + 1, n - 1):
            visited = set()
            values = [0, 0, 0]  # To store the XOR values of the three components
            edge1 = edges[i]
            edge2 = edges[j]

            # Remove the edges temporarily
            graph[edge1[0]].remove(edge1[1])
            graph[edge1[1]].remove(edge1[0])
            graph[edge2[0]].remove(edge2[1])
            graph[edge2[1]].remove(edge2[0])

            # Perform DFS to get the three components
            for node in range(n):
                if node not in visited:
                    dfs(node, visited, [values[len(visited) // (n // 3)]])
            
            # Restore the edges
            graph[edge1[0]].append(edge1[1])
            graph[edge1[1]].append(edge1[0])
            graph[edge2[0]].append(edge2[1])
            graph[edge2[1]].append(edge2[0])

            # Calculate score
            score = max(values) - min(values)
            min_score = min(min_score, score)

    return min_score
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