How do you solve Count Nodes With the Highest Score on LeetCode?

Count Nodes With the Highest Score (LeetCode #2049) 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 how to calculate subtree sizes is crucial for optimizing the score calculation.

What is the brute force solution for Count Nodes With the Highest Score?

The brute force approach for Count Nodes With the Highest Score is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach calculates the score for each node by removing it and counting the sizes of the resulting subtrees. This is straightforward but inefficient, as it requires recalculating subtree sizes multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Nodes With the Highest Score?

Count Nodes With the Highest Score uses the following concepts: Array, Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Tree Traversal.

What are the interview tips for Count Nodes With the Highest Score?

Explain your thought process clearly while coding; it shows your understanding. Consider edge cases, such as the smallest trees, to ensure your solution is robust. Practice DFS and tree traversal problems to become comfortable with recursive approaches.

What are common mistakes when solving Count Nodes With the Highest Score?

Failing to correctly handle the case when the node is the root (node 0). Not properly initializing data structures to store subtree sizes.

#2049

Count Nodes With the Highest Score

Medium

ArrayTreeDepth-First SearchBinary TreeDepth-First SearchTree Traversal

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 uses a single DFS to calculate subtree sizes and scores in one pass, reducing redundant calculations. By leveraging the total number of nodes, we can efficiently compute scores without recalculating subtree sizes multiple times.

⚙️

Algorithm

3 steps

1Step 1: Perform a DFS to calculate the size of each subtree and store it.
2Step 2: Use the total size of the tree to compute the score for each node based on its children's subtree sizes.
3Step 3: Track the maximum score and count how many nodes achieve this score.

solution.py27 lines

1def countHighestScoreNodes(parents):
2    n = len(parents)
3    children = [[] for _ in range(n)]
4    for i in range(1, n):
5        children[parents[i]].append(i)
6    subtree_sizes = [0] * n
7    def dfs(node):
8        size = 1
9        for child in children[node]:
10            size += dfs(child)
11        subtree_sizes[node] = size
12        return size
13    dfs(0)
14    max_score = 0
15    count = 0
16    for i in range(n):
17        score = 1
18        for child in children[i]:
19            score *= subtree_sizes[child]
20        if i != 0:
21            score *= (n - subtree_sizes[i])
22        if score > max_score:
23            max_score = score
24            count = 1
25        elif score == max_score:
26            count += 1
27    return count

ℹ

Complexity note: This complexity is efficient because we only traverse the tree once to calculate subtree sizes and then again to compute scores, leading to a linear time complexity.

1Understanding how to calculate subtree sizes is crucial for optimizing the score calculation.
2Using DFS allows us to gather necessary information in a single pass, avoiding redundant calculations.

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