Leetcode #2492: Minimum Score of a Path Between Two Cities
In this guide, we solve Leetcode #2492 Minimum Score of a Path Between Two Cities 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 positive integer n representing n cities numbered from 1 to n. You are also given a 2D array roads where roads[i] = [ai, bi, distancei] indicates that there is a bidirectional road between cities ai and bi with a distance equal to distancei.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Union Find, Graph
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
Input: n = 4, roads = [[1,2,9],[2,3,6],[2,4,5],[1,4,7]]
Output: 5
Explanation: The path from city 1 to 4 with the minimum score is: 1 -> 2 -> 4. The score of this path is min(9,5) = 5.
It can be shown that no other path has less score.
Python Solution
class Solution:
def minScore(self, n: int, roads: List[List[int]]) -> int:
def dfs(i):
nonlocal ans
for j, d in g[i]:
ans = min(ans, d)
if not vis[j]:
vis[j] = True
dfs(j)
g = defaultdict(list)
for a, b, d in roads:
g[a].append((b, d))
g[b].append((a, d))
vis = [False] * (n + 1)
ans = inf
dfs(1)
return ans
Complexity
The time complexity is , where and are the number of nodes and edges, respectively. The space complexity is O(V).
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.