Leetcode #2065: Maximum Path Quality of a Graph
In this guide, we solve Leetcode #2065 Maximum Path Quality of a Graph 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
There is an undirected graph with n nodes numbered from 0 to n - 1 (inclusive). You are given a 0-indexed integer array values where values[i] is the value of the ith node.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Graph, Array, Backtracking
Intuition
We must explore combinations of choices, but many branches can be pruned early.
Backtracking enumerates valid candidates while keeping the search space under control.
Approach
Use DFS to build candidates step by step, and backtrack when constraints are violated.
Pruning keeps the exploration practical for typical constraints.
Steps:
- Define the decision tree.
- DFS through choices and backtrack.
- Prune invalid paths early.
Example
Input: values = [0,32,10,43], edges = [[0,1,10],[1,2,15],[0,3,10]], maxTime = 49
Output: 75
Explanation:
One possible path is 0 -> 1 -> 0 -> 3 -> 0. The total time taken is 10 + 10 + 10 + 10 = 40 <= 49.
The nodes visited are 0, 1, and 3, giving a maximal path quality of 0 + 32 + 43 = 75.
Python Solution
class Solution:
def maximalPathQuality(
self, values: List[int], edges: List[List[int]], maxTime: int
) -> int:
def dfs(u: int, cost: int, value: int):
if u == 0:
nonlocal ans
ans = max(ans, value)
for v, t in g[u]:
if cost + t <= maxTime:
if vis[v]:
dfs(v, cost + t, value)
else:
vis[v] = True
dfs(v, cost + t, value + values[v])
vis[v] = False
n = len(values)
g = [[] for _ in range(n)]
for u, v, t in edges:
g[u].append((v, t))
g[v].append((u, t))
vis = [False] * n
vis[0] = True
ans = 0
dfs(0, 0, values[0])
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.