Solution Explanation: Counting Nodes with the Highest Score

This problem involves traversing a tree to calculate scores for each node and then finding the number of nodes with the highest score. The score of a node is calculated by removing the node and its connecting edges, resulting in one or more subtrees. The score is the product of the sizes of these subtrees.

Algorithm:

1. Tree Construction: The input parents array represents the tree structure. We build an adjacency list (graph) g to efficiently access children of each node.

2. Depth-First Search (DFS): A recursive DFS function dfs(i, fa) is the core of the solution. It takes the current node i and its parent fa as input.

3. Score Calculation within DFS:

   • cnt: Tracks the size of the subtree rooted at the current node i. It's initialized to 1 (the node itself).
   • score: Keeps track of the product of subtree sizes. It's initialized to 1.
   • The DFS recursively explores the children of i. For each child j (excluding the parent fa to avoid cycles), it recursively calls dfs(j, i).
   • The return value of the recursive call (t) represents the size of the subtree rooted at child j. This t is multiplied into the score and added to cnt.
   • After processing all children, if there are remaining nodes in the original tree (n - cnt > 0), this represents a subtree resulting from removing node i, and its size is multiplied into score.

4. Tracking Highest Score:

   • mx: Stores the current highest score encountered.
   • ans: Counts the number of nodes with the highest score.
   • After calculating score for node i, it's compared with mx. If score > mx, mx is updated, and ans is set to 1. If score == mx, ans is incremented.

5. Return Value: The function returns ans, which is the number of nodes with the highest score.

Time Complexity: O(N), where N is the number of nodes. The DFS visits each node exactly once.

Space Complexity: O(N). The space is dominated by the adjacency list g and the recursive call stack during DFS. In the worst case (a skewed tree), the stack depth can be O(N).

Code Examples (Python):

class Solution:
    def countHighestScoreNodes(self, parents: List[int]) -> int:
        n = len(parents)
        g = [[] for _ in range(n)]
        for i in range(1, n):
            g[parents[i]].append(i)
        ans = mx = 0
 
        def dfs(i, fa):
            nonlocal ans, mx
            cnt = score = 1
            for j in g[i]:
                if j != fa:
                    t = dfs(j, i)
                    score *= t
                    cnt += t
            if n - cnt > 0:
                score *= n - cnt
            if mx < score:
                mx = score
                ans = 1
            elif mx == score:
                ans += 1
            return cnt
 
        dfs(0, -1)
        return ans
 

The code in other languages follows the same logic, with minor syntactic differences. The provided code examples in Java, C++, Go, TypeScript, and C# all implement this efficient DFS-based approach.