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:
[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.
Example 1:
Input: nums = [1,5,5,4,11], edges = [[0,1],[1,2],[1,3],[3,4]] Output: 9 Explanation: The diagram above shows a way to make a pair of removals. - The 1st component has nodes [1,3,4] with values [5,4,11]. Its XOR value is 5 ^ 4 ^ 11 = 10. - The 2nd component has node [0] with value [1]. Its XOR value is 1 = 1. - The 3rd component has node [2] with value [5]. Its XOR value is 5 = 5. The score is the difference between the largest and smallest XOR value which is 10 - 1 = 9. It can be shown that no other pair of removals will obtain a smaller score than 9.
Example 2:
Input: nums = [5,5,2,4,4,2], edges = [[0,1],[1,2],[5,2],[4,3],[1,3]] Output: 0 Explanation: The diagram above shows a way to make a pair of removals. - The 1st component has nodes [3,4] with values [4,4]. Its XOR value is 4 ^ 4 = 0. - The 2nd component has nodes [1,0] with values [5,5]. Its XOR value is 5 ^ 5 = 0. - The 3rd component has nodes [2,5] with values [2,2]. Its XOR value is 2 ^ 2 = 0. The score is the difference between the largest and smallest XOR value which is 0 - 0 = 0. We cannot obtain a smaller score than 0.
Constraints:
n == nums.length3 <= n <= 10001 <= nums[i] <= 108edges.length == n - 1edges[i].length == 20 <= ai, bi < nai != biedges represents a valid tree.The key idea is to perform a depth-first search (DFS) to compute the XOR for each subtree rooted at each node. By caching these values, we can efficiently determine the XOR of components after removing two edges.
We can use these cached values to compute the possible component XOR values after removing each pair of edges, and calculate the score of each pair.
This Python solution first builds an adjacency list from the given edges. It then uses DFS from an arbitrary root (node 0) to compute the XOR of every subtree rooted at each node. It utilizes this subtree data to compute possible XOR values when two edges are removed, and tracks the minimum score obtained.
The use of sets and filtering ensures efficient edge selection to deeply examine the implied subtrees created by removing distinct pairs of edges.
JavaScript
Time Complexity: O(n^2), where n is the number of nodes, due to the nested loops for examining edge pairs.
Space Complexity: O(n), mainly consumed by the adjacency list and XOR storage.
This approach leverages a brute-force analysis, with optimizations such as subtree XOR precomputation and efficient edge double-exclusion considerations.
The key is to evaluate every pair of distinct edges, ensuring each removal leads to valid and distinct subcomponent calculations.
The C++ solution optimizes the brute-force method to reduce redundant calculations when determining unique edge removal pairs. This code calculates all subtree XOR combinations and evaluates the score for each distinct pair of potential edge removals.
Adjacency lists and careful nested loop indices ensure complete yet efficient configurations of potential subtree analysis.
Java
Time Complexity: O(n^3) due to nested loops checking each edge in combination with others.
Space Complexity: O(n), using vectors to manage adjacency and subtree XORs.
| Approach | Complexity |
|---|---|
| DFS with Subtree XOR Caching | Time Complexity: O(n^2), where n is the number of nodes, due to the nested loops for examining edge pairs. Space Complexity: O(n), mainly consumed by the adjacency list and XOR storage. |
| Brute Force with Optimizations | Time Complexity: O(n^3) due to nested loops checking each edge in combination with others. Space Complexity: O(n), using vectors to manage adjacency and subtree XORs. |
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