Introduction

In this blog post, we will explore how to solve the LeetCode problem "2049" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

There is a binary tree rooted at 0 consisting of n nodes. The nodes are labeled from 0 to n - 1. You are given a 0-indexed integer array parents representing the tree, where parents[i] is the parent of node i. Since node 0 is the root, parents[0] == -1. Each node has a score. To find the score of a node, consider if the node and the edges connected to it were removed. The tree would become one or more non-empty subtrees. The size of a subtree is the number of the nodes in it. The score of the node is the product of the sizes of all those subtrees. Return the number of nodes that have the highest score. Example 1: Input: parents = [-1,2,0,2,0] Output: 3 Explanation: - The score of node 0 is: 3 1 = 3 - The score of node 1 is: 4 = 4 - The score of node 2 is: 1 1 2 = 2 - The score of node 3 is: 4 = 4 - The score of node 4 is: 4 = 4 The highest score is 4, and three nodes (node 1, node 3, and node 4) have the highest score. Example 2: Input: parents = [-1,2,0] Output: 2 Explanation: - The score of node 0 is: 2 = 2 - The score of node 1 is: 2 = 2 - The score of node 2 is: 1 1 = 1 The highest score is 2, and two nodes (node 0 and node 1) have the highest score. Constraints: n == parents.length 2 <= n <= 105 parents[0] == -1 0 <= parents[i] <= n - 1 for i != 0 parents represents a valid binary tree.

Explanation

Here's the approach and Python code to solve the problem efficiently:

• Build Adjacency List: Create an adjacency list representation of the binary tree from the parents array. This allows us to easily find the children of each node.
• Calculate Subtree Sizes: Perform a Depth-First Search (DFS) to calculate the size of each subtree rooted at each node. This information is crucial for computing the score of each node.

• Calculate Scores and Track Max: Iterate through each node, calculate its score based on the subtree sizes, and keep track of the highest score encountered so far and the number of nodes with that score.

• Runtime Complexity: O(N), where N is the number of nodes in the tree.

• Storage Complexity: O(N) due to the adjacency list and subtree size array.

Code

    def countHighestScoreNodes(parents):
    n = len(parents)
    children = [[] for _ in range(n)]
    for i in range(1, n):
        children[parents[i]].append(i)

    subtree_sizes = [0] * n

    def dfs(node):
        subtree_sizes[node] = 1
        for child in children[node]:
            subtree_sizes[node] += dfs(child)
        return subtree_sizes[node]

    dfs(0)

    max_score = 0
    count = 0

    for i in range(n):
        score = 1
        num_children = len(children[i])

        if num_children == 0:
            score = n - 1
        elif num_children == 1:
            score = subtree_sizes[children[i][0]]
            if n - subtree_sizes[i] > 0:
                score *= (n - subtree_sizes[i])
        else:
            score = subtree_sizes[children[i][0]] * subtree_sizes[children[i][1]]
            if n - subtree_sizes[i] > 0:
                score *= (n - subtree_sizes[i])

        if score > max_score:
            max_score = score
            count = 1
        elif score == max_score:
            count += 1

    return count