从树中删除边的最小分数

/*
 * 思路：
 * 1. 使用流操作简化异或值计算和分数计算。
 * 2. 遍历所有可能的删除边组合。
 * 3. 使用流操作处理图结构。
 */
import java.util.*;
import java.util.stream.Collectors;

public class Solution {
    public int minimumScore(int[] nums, int[][] edges) {
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            graph.add(new ArrayList<>());
        }
        Arrays.stream(edges).forEach(edge -> {
            graph.get(edge[0]).add(edge[1]);
            graph.get(edge[1]).add(edge[0]);
        });
        int[] xor = new int[nums.length];
        boolean[] visited = new boolean[nums.length];
        dfs(0, -1, graph, nums, xor, visited);
        return Arrays.stream(edges)
                     .flatMapToInt(edge1 -> Arrays.stream(edges).filter(edge2 -> !Arrays.equals(edge1, edge2)).mapToInt(edge2 -> calculateScore(xor, edge1, edge2, graph, nums)))
                     .min()
                     .orElse(Integer.MAX_VALUE);
    }

    private void dfs(int node, int parent, List<List<Integer>> graph, int[] nums, int[] xor, boolean[] visited) {
        visited[node] = true;
        xor[node] = nums[node];
        for (int neighbor : graph.get(node)) {
            if (neighbor != parent && !visited[neighbor]) {
                dfs(neighbor, node, graph, nums, xor, visited);
                xor[node] ^= xor[neighbor];
            }
        }
    }

    private int calculateScore(int[] xor, int[] edge1, int[] edge2, List<List<Integer>> graph, int[] nums) {
        int node1 = edge1[0], node2 = edge1[1];
        int node3 = edge2[0], node4 = edge2[1];
        int xor1 = xor[node1], xor2 = xor[node2];
        int xor3 = xor[node3], xor4 = xor[node4];
        int[] components = {xor1, xor2, xor3, xor4};
        Arrays.sort(components);
        return components[3] - components[0];
    }
}