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2492. 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.

Solutions

Solution 1: DFS

According to the problem description, each edge can be passed multiple times, and it is guaranteed that node $1$ and node $n$ are in the same connected component. Therefore, the problem is actually looking for the smallest edge in the connected component where node $1$ is located. We can use DFS, start searching from node $1$, and find the smallest edge.

The time complexity is $O(n + m)$, where $n$ and $m$ are the number of nodes and edges, respectively.

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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

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class Solution {
    private List<int[]>[] g;
    private boolean[] vis;
    private int ans = 1 << 30;

    public int minScore(int n, int[][] roads) {
        g = new List[n];
        vis = new boolean[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : roads) {
            int a = e[0] - 1, b = e[1] - 1, d = e[2];
            g[a].add(new int[] {b, d});
            g[b].add(new int[] {a, d});
        }
        dfs(0);
        return ans;
    }

    private void dfs(int i) {
        for (var nxt : g[i]) {
            int j = nxt[0], d = nxt[1];
            ans = Math.min(ans, d);
            if (!vis[j]) {
                vis[j] = true;
                dfs(j);
            }
        }
    }
}

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class Solution {
public:
    int minScore(int n, vector<vector<int>>& roads) {
        vector<vector<pair<int, int>>> g(n);
        bool vis[n];
        memset(vis, 0, sizeof vis);
        for (auto& e : roads) {
            int a = e[0] - 1, b = e[1] - 1, d = e[2];
            g[a].emplace_back(b, d