Leetcode #2374: Node With Highest Edge Score
In this guide, we solve Leetcode #2374 Node With Highest Edge Score in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a directed graph with n nodes labeled from 0 to n - 1, where each node has exactly one outgoing edge. The graph is represented by a given 0-indexed integer array edges of length n, where edges[i] indicates that there is a directed edge from node i to node edges[i].
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Graph, Hash Table
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: edges = [1,0,0,0,0,7,7,5]
Output: 7
Explanation:
- The nodes 1, 2, 3 and 4 have an edge pointing to node 0. The edge score of node 0 is 1 + 2 + 3 + 4 = 10.
- The node 0 has an edge pointing to node 1. The edge score of node 1 is 0.
- The node 7 has an edge pointing to node 5. The edge score of node 5 is 7.
- The nodes 5 and 6 have an edge pointing to node 7. The edge score of node 7 is 5 + 6 = 11.
Node 7 has the highest edge score so return 7.
Python Solution
class Solution:
def edgeScore(self, edges: List[int]) -> int:
ans = 0
cnt = [0] * len(edges)
for i, j in enumerate(edges):
cnt[j] += i
if cnt[ans] < cnt[j] or (cnt[ans] == cnt[j] and j < ans):
ans = j
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.