- Difficulty: Hard
- Tags: LeetCode, Hard, Bit Manipulation, Tree, Depth-First Search, Array, leetcode-2322, O(n^2), O(n), DFS
Problem
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:
- 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 are4 ^ 5 ^ 7 = 6
,1 ^ 9 = 8
, and3 ^ 3 ^ 3 = 3
. The largest XOR value is8
and the smallest XOR value is3
. The score is then8 - 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.length
3 <= n <= 1000
1 <= nums[i] <= 108
edges.length == n - 1
edges[i].length == 2
0 <= ai, bi < n
ai != bi
edges
represents a valid tree.