LeetCode-in-Java

3112. Minimum Time to Visit Disappearing Nodes

Medium

There is an undirected graph of n nodes. You are given a 2D array edges, where edges[i] = [ui, vi, lengthi] describes an edge between node ui and node vi with a traversal time of lengthi units.

Additionally, you are given an array disappear, where disappear[i] denotes the time when the node i disappears from the graph and you won’t be able to visit it.

Notice that the graph might be disconnected and might contain multiple edges.

Return the array answer, with answer[i] denoting the minimum units of time required to reach node i from node 0. If node i is unreachable from node 0 then answer[i] is -1.

Example 1:

Input: n = 3, edges = [[0,1,2],[1,2,1],[0,2,4]], disappear = [1,1,5]

Output: [0,-1,4]

Explanation:

We are starting our journey from node 0, and our goal is to find the minimum time required to reach each node before it disappears.

• For node 0, we don’t need any time as it is our starting point.
• For node 1, we need at least 2 units of time to traverse edges[0]. Unfortunately, it disappears at that moment, so we won’t be able to visit it.
• For node 2, we need at least 4 units of time to traverse edges[2].

Example 2:

Input: n = 3, edges = [[0,1,2],[1,2,1],[0,2,4]], disappear = [1,3,5]

Output: [0,2,3]

Explanation:

We are starting our journey from node 0, and our goal is to find the minimum time required to reach each node before it disappears.

• For node 0, we don’t need any time as it is the starting point.
• For node 1, we need at least 2 units of time to traverse edges[0].
• For node 2, we need at least 3 units of time to traverse edges[0] and edges[1].

Example 3:

Input: n = 2, edges = [[0,1,1]], disappear = [1,1]

Output: [0,-1]

Explanation:

Exactly when we reach node 1, it disappears.

Constraints:

• 1 <= n <= 5 * 104
• 0 <= edges.length <= 105
• edges[i] == [ui, vi, lengthi]
• 0 <= ui, vi <= n - 1
• 1 <= lengthi <= 105
• disappear.length == n
• 1 <= disappear[i] <= 105

Solution

import java.util.Arrays;

public class Solution {
    public int[] minimumTime(int n, int[][] edges, int[] disappear) {
        int[] dist = new int[n];
        Arrays.fill(dist, Integer.MAX_VALUE);
        boolean exit = false;
        int i;
        int src;
        int dest;
        int cost;
        dist[0] = 0;
        for (i = 0; i < n && !exit; ++i) {
            exit = true;
            for (int[] edge : edges) {
                src = edge[0];
                dest = edge[1];
                cost = edge[2];
                if (dist[src] != -1
                        && dist[src] != Integer.MAX_VALUE
                        && dist[src] < disappear[src]
                        && dist[src] + cost < dist[dest]) {
                    exit = false;
                    dist[dest] = dist[src] + cost;
                }
                if (dist[dest] != -1
                        && dist[dest] != Integer.MAX_VALUE
                        && dist[dest] < disappear[dest]
                        && dist[dest] + cost < dist[src]) {
                    exit = false;
                    dist[src] = dist[dest] + cost;
                }
            }
        }
        for (i = 0; i < dist.length; ++i) {
            if (dist[i] == Integer.MAX_VALUE || dist[i] >= disappear[i]) {
                dist[i] = -1;
            }
        }
        return dist;
    }
}

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