Minimum Score After Removals on a Tree

This problem involves finding the minimum score achievable by removing two edges from a tree, resulting in three connected components. The score is calculated as the difference between the largest and smallest XOR sums of the values in each component.

Approach

The solution uses Depth-First Search (DFS) to efficiently calculate XOR sums of components. The overall strategy is as follows:

1. Preprocessing: Build an adjacency list representation of the tree from the given edges. Calculate the XOR sum (s) of all node values in the tree.

2. Iterate through Edge Pairs: The algorithm iterates through all possible pairs of edges to remove. For each pair:

3. DFS to Calculate XOR Sums: After removing a pair of edges, we are left with three connected components. Two nested DFS functions are used:

   • dfs(i, fa, x): Calculates the XOR sum of a subtree rooted at node i, excluding the subtree rooted at node x and avoiding cycles by tracking the parent node fa. This is crucial to isolating the components.
   • dfs2(i, fa, x): This DFS function recursively explores the tree, calculating the XOR sum of the current component and updates the minimum score (ans).

4. Calculate and Update Score: For each component, the XOR sum is computed using dfs. The difference between the largest and smallest XOR sums among the three components is the score for the current edge pair. The minimum score found so far is updated.

5. Return Minimum Score: After exploring all edge pairs, the function returns the minimum score found.

Time Complexity Analysis

• Building the adjacency list takes O(E), where E is the number of edges (E = N-1 for a tree).
• Iterating through edge pairs takes O(E^2) = O(N^2) time.
• Each DFS call takes O(N) in the worst case.
• Therefore, the overall time complexity is dominated by the nested loops and DFS calls, resulting in O(N^3).

Space Complexity Analysis

• The adjacency list uses O(E) = O(N) space.
• The recursive DFS calls use space proportional to the recursion depth, which is O(N) in the worst case (a completely skewed tree).
• Therefore, the overall space complexity is O(N).

Code Explanation (Python)

class Solution:
    def minimumScore(self, nums: List[int], edges: List[List[int]]) -> int:
        def dfs(i, fa, x): #DFS to calculate XOR sum of a component
            res = nums[i]
            for j in g[i]:
                if j != fa and j != x:
                    res ^= dfs(j, i, x)
            return res
 
        def dfs2(i, fa, x): #Recursive DFS to explore components and update min score
            nonlocal s, s1, ans
            res = nums[i]
            for j in g[i]:
                if j != fa and j != x:
                    a = dfs2(j, i, x)
                    res ^= a
                    b = s1 ^ a
                    c = s ^ s1
                    t = max(a, b, c) - min(a, b, c)
                    ans = min(ans, t)
            return res
 
        g = defaultdict(list) #Adjacency list
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
 
        s = 0 #XOR sum of all nodes
        for v in nums:
            s ^= v
        n = len(nums)
        ans = inf #Initialize minimum score to infinity
        for i in range(n): #Iterate through edge pairs
            for j in g[i]:
                s1 = dfs(i, -1, j) #Calculate XOR sum of first component
                dfs2(i, -1, j) #Recursive DFS to explore other components and update score
        return ans

The code in other languages (Java, C++, Go) follows a similar structure and logic, with appropriate syntax changes for each language. The core algorithms remain the same: building the adjacency list, iterating through edge pairs, performing DFS to compute component XOR sums, and updating the minimum score.