Leetcode #2322: Minimum Score After Removals on a Tree
In this guide, we solve Leetcode #2322 Minimum Score After Removals on a Tree in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Tree, Depth-First Search, Array
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
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.
Python Solution
class Solution:
def minimumScore(self, nums: List[int], edges: List[List[int]]) -> int:
def dfs(i: int, fa: int) -> int:
res = nums[i]
for j in g[i]:
if j != fa:
res ^= dfs(j, i)
return res
def dfs2(i: int, fa: int) -> int:
nonlocal s, s1, ans
res = nums[i]
for j in g[i]:
if j != fa:
s2 = dfs2(j, i)
res ^= s2
mx = max(s ^ s1, s2, s1 ^ s2)
mn = min(s ^ s1, s2, s1 ^ s2)
ans = min(ans, mx - mn)
return res
g = defaultdict(list)
for a, b in edges:
g[a].append(b)
g[b].append(a)
s = reduce(lambda x, y: x ^ y, nums)
n = len(nums)
ans = inf
for i in range(n):
for j in g[i]:
s1 = dfs(i, j)
dfs2(i, j)
return ans
Complexity
The time complexity is , and the space complexity is , where is the number of nodes in the tree. The space complexity is , where is the number of nodes in the tree.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.