2737. Find the Closest Marked Node 🔒

Description

You are given a positive integer n which is the number of nodes of a 0-indexed directed weighted graph and a 0-indexed 2D array edges where edges[i] = [u_i, v_i, w_i] indicates that there is an edge from node u_i to node v_i with weight w_i.

You are also given a node s and a node array marked; your task is to find the minimum distance from s to any of the nodes in marked.

Return an integer denoting the minimum distance from s to any node in marked or -1 if there are no paths from s to any of the marked nodes.

Example 1:

Input: n = 4, edges = [[0,1,1],[1,2,3],[2,3,2],[0,3,4]], s = 0, marked = [2,3]
Output: 4
Explanation: There is one path from node 0 (the green node) to node 2 (a red node), which is 0->1->2, and has a distance of 1 + 3 = 4.
There are two paths from node 0 to node 3 (a red node), which are 0->1->2->3 and 0->3, the first one has a distance of 1 + 3 + 2 = 6 and the second one has a distance of 4.
The minimum of them is 4.

Example 2:

Input: n = 5, edges = [[0,1,2],[0,2,4],[1,3,1],[2,3,3],[3,4,2]], s = 1, marked = [0,4]
Output: 3
Explanation: There are no paths from node 1 (the green node) to node 0 (a red node).
There is one path from node 1 to node 4 (a red node), which is 1->3->4, and has a distance of 1 + 2 = 3.
So the answer is 3.

Example 3:

Input: n = 4, edges = [[0,1,1],[1,2,3],[2,3,2]], s = 3, marked = [0,1]
Output: -1
Explanation: There are no paths from node 3 (the green node) to any of the marked nodes (the red nodes), so the answer is -1.

Constraints:

2 <= n <= 500
1 <= edges.length <= 10⁴
edges[i].length = 3
0 <= edges[i][0], edges[i][1] <= n - 1
1 <= edges[i][2] <= 10⁶
1 <= marked.length <= n - 1
0 <= s, marked[i] <= n - 1
s != marked[i]
marked[i] != marked[j] for every i != j
The graph might have repeated edges.
The graph is generated such that it has no self-loops.

Solutions

Solution 1: Dijkstra's Algorithm

First, we construct an adjacency matrix $g$ based on the edge information provided in the problem, where $g[i][j]$ represents the distance from node $i$ to node $j$. If such an edge does not exist, then $g[i][j]$ is positive infinity.

Then, we can use Dijkstra's algorithm to find the shortest distance from the starting point $s$ to all nodes, denoted as $dist$.

Finally, we traverse all the marked nodes and find the marked node with the smallest distance. If the distance is positive infinity, we return $-1$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the number of nodes.

Python3JavaC++GoTypeScript

class Solution:
    def minimumDistance(
        self, n: int, edges: List[List[int]], s: int, marked: List[int]
    ) -> int:
        g = [[inf] * n for _ in range(n)]
        for u, v, w in edges:
            g[u][v] = min(g[u][v], w)
        dist = [inf] * n
        vis = [False] * n
        dist[s] = 0
        for _ in range(n):
            t = -1
            for j in range(n):
                if not vis[j] and (t == -1 or dist[t] > dist[j]):
                    t = j
            vis[t] = True
            for j in range(n):
                dist[j] = min(dist[j], dist[t] + g[t][j])
        ans = min(dist[i] for i in marked)
        return -1 if ans >= inf else ans

class Solution {
    public int minimumDistance(int n, List<List<Integer>> edges, int s, int[] marked) {
        final int inf = 1 << 29;
        int[][] g = new int[n][n];
        for (var e : g) {
            Arrays.fill(e, inf);
        }
        for (var e : edges) {
            int u = e.get(0), v = e.get(1), w = e.get(2);
            g[u][v] = Math.min(g[u][v], w);
        }
        int[] dist = new int[n];
        Arrays.fill(dist, inf);
        dist[s] = 0;
        boolean[] vis = new boolean[n];
        for (int i = 0; i < n; ++i) {
            int t = -1;
            for (int j = 0; j < n; ++j) {
                if (!vis[j] && (t == -1 || dist[t] > dist[j])) {
                    t = j;
                }
            }
            vis[t] = true;
            for (int j = 0; j < n; ++j) {
                dist[j] = Math.min(dist[j], dist[t] + g[t][j]);
            }
        }
        int ans = inf;
        for (int i : marked) {
            ans = Math.min(ans, dist[i]);
        }
        return ans >= inf ? -1 : ans;
    }
}

class Solution 