# https://leetcode.com/problems/minimum-score-of-a-path-between-two-cities/description/
"""
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.

Constraints:
2 <= n <= 105
1 <= roads.length <= 105
roads[i].length == 3
1 <= ai, bi <= n
ai != bi
1 <= distancei <= 104
There are no repeated edges.
There is at least one path between 1 and n.
"""
from math import inf

from puzzles.union_find import UF


def min_score(n: int, roads: list[list[int]]) -> int:
    min_score_map = [inf] * (n + 1)
    uf = UF(n + 1)
    for u, v, score in roads:
        uf.union(u, v)
        min_score_map[u] = min(min_score_map[u], score)
        min_score_map[v] = min(min_score_map[v], score)
    return min((min_score_map[i] for i in range(1, n + 1) if uf.connected(i, 1)))
