How do you solve Minimum Score After Removals on a Tree on LeetCode?

Minimum Score After Removals on a Tree (LeetCode #2322) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the properties of XOR can significantly reduce the complexity of the problem.

What is the brute force solution for Minimum Score After Removals on a Tree?

The brute force approach for Minimum Score After Removals on a Tree is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible pair of edges to remove and calculating the resulting XOR values for the three components. This method is straightforward but inefficient, as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Score After Removals on a Tree?

Minimum Score After Removals on a Tree uses the following concepts: Array, Bit Manipulation, Tree, Depth-First Search. The recommended patterns to study are: Tree Traversal, Dynamic Programming on Trees.

What are the interview tips for Minimum Score After Removals on a Tree?

Always clarify the problem requirements and constraints before diving into the solution. Consider edge cases and how they might affect your approach. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Score After Removals on a Tree?

Failing to account for all components after edge removals. Not leveraging the properties of XOR, leading to unnecessary recalculations.

#2322

Minimum Score After Removals on a Tree

Hard

ArrayBit ManipulationTreeDepth-First SearchTree TraversalDynamic Programming on Trees

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach leverages the properties of XOR and tree structure. By calculating the XOR for each subtree in advance, we can efficiently determine the scores for different edge removals without redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total XOR of all node values.
2Step 2: For each edge, calculate the XOR of the subtree formed by removing that edge.
3Step 3: Store the XOR values of all subtrees. For each pair of edges, compute the score using the precomputed values.

solution.py36 lines

1# Full working Python code
2import collections
3
4class Solution:
5    def minScore(self, nums, edges):
6        graph = collections.defaultdict(list)
7        for a, b in edges:
8            graph[a].append(b)
9            graph[b].append(a)
10
11        total_xor = 0
12        for num in nums:
13            total_xor ^= num
14
15        subtree_xor = [0] * len(nums)
16        visited = [False] * len(nums)
17
18        def dfs(node):
19            visited[node] = True
20            xor_sum = nums[node]
21            for neighbor in graph[node]:
22                if not visited[neighbor]:
23                    xor_sum ^= dfs(neighbor)
24            subtree_xor[node] = xor_sum
25            return xor_sum
26
27        dfs(0)
28
29        min_score = float('inf')
30        for a, b in edges:
31            xor1 = subtree_xor[a]
32            xor2 = total_xor ^ xor1
33            score = abs(xor1 - xor2)
34            min_score = min(min_score, score)
35
36        return min_score

ℹ

Complexity note: This complexity arises because we traverse the tree once to compute the total XOR and subtree XORs, making it efficient compared to the brute force method.

1Understanding the properties of XOR can significantly reduce the complexity of the problem.
2Precomputing values like subtree XORs allows for efficient calculations when considering edge removals.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.