# 2492. Minimum Score of a Path Between Two Cities
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. The cities graph is not necessarily connected.
The score of a path between two cities is defined as the minimum distance of a road in this path.
Return the minimum possible score of a path between cities 1 and n.
Note:
A path is a sequence of roads between two cities.
It is allowed for a path to contain the same road multiple times, and you can visit cities 1 and n multiple times along the path.
The test cases are generated such that there is at least one path between 1 and n.
Example 1:
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.
Example 2:
Input: n = 4, roads = [[1,2,2],[1,3,4],[3,4,7]]
Output: 2
Explanation: The path from city 1 to 4 with the minimum score is: 1 -> 2 -> 1 -> 3 -> 4. The score of this path is min(2,2,4,7) = 2.
# Solution
Approach 1: just find that min edge in the connected component containing node 1.
# Code (Python)
Approach 1:
from collections import defaultdict
class Solution:
def minScore(self, n: int, roads: list[list[int]]) -> int:
# just find that min edge in the connected component containing node 1
edges = defaultdict(list)
for end1, end2, value in roads:
edges[end1].append((end2, value))
edges[end2].append((end1, value))
min_edge = [float('inf')]
self._recurse(1, edges, set(), min_edge)
return min_edge[0]
def _recurse(self, node, edges, seen, min_edge):
if node in seen:
return
seen.add(node)
for neighbor, value in edges[node]:
min_edge[0] = min(min_edge[0], value)
self._recurse(neighbor, edges, seen, min_edge)
Approach 2:
# Code (C++)
Approach 1:
Approach 2: