How do you solve Node With Highest Edge Score on LeetCode?

Node With Highest Edge Score (LeetCode #2374) 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: Each node has exactly one outgoing edge, simplifying the score calculation.

What is the brute force solution for Node With Highest Edge Score?

The brute force approach for Node With Highest Edge Score is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we calculate the edge score for each node by iterating through all nodes and summing the indices of nodes that point to it. This is straightforward but inefficient for larger graphs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Node With Highest Edge Score?

Node With Highest Edge Score uses the following concepts: Hash Table, Graph Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Node With Highest Edge Score?

Clearly explain your thought process while coding. Consider edge cases, such as multiple nodes with the same score. Optimize your solution after the brute force approach.

What are common mistakes when solving Node With Highest Edge Score?

Not considering the smallest index in case of ties. Misunderstanding how to accumulate scores for nodes.

#2374

Node With Highest Edge Score

Medium

Hash TableGraph TheoryHash MapArray

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)

In the optimal approach, we use a single pass to calculate the edge scores by leveraging the fact that each node has exactly one outgoing edge. This allows us to efficiently compute scores in O(n) time.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array `score` of size n with all zeros to store the edge scores for each node.
2Step 2: Iterate through the `edges` array. For each index `i`, add `i` to `score[edges[i]]` since `edges[i]` is the node that `i` points to.
3Step 3: After populating the scores, iterate through the `score` array to find the node with the maximum score, ensuring to return the smallest index in case of ties.

solution.py14 lines

1# Full working Python code
2
3def highestEdgeScore(edges):
4    n = len(edges)
5    score = [0] * n
6    for i in range(n):
7        score[edges[i]] += i
8    max_score = -1
9    result_node = -1
10    for i in range(n):
11        if score[i] > max_score or (score[i] == max_score and i < result_node):
12            max_score = score[i]
13            result_node = i
14    return result_node

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the edges to calculate scores and another pass to find the maximum. The space complexity is O(n) due to the score array used to store edge scores.

1Each node has exactly one outgoing edge, simplifying the score calculation.
2The edge score is cumulative based on the indices of nodes pointing to it.

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