#3112

Medium

LeetCode Problem Workspace

Minimum Time to Visit Disappearing Nodes

Determine the minimum time to visit each node in a disappearing-node graph using arrays, graphs, and priority queues efficiently.

array graph heap priority queue shortest path

Problem Statement

You are given an undirected graph of n nodes labeled from 0 to n-1. The graph is defined by a list of edges, where edges[i] = [ui, vi, lengthi] represents a bidirectional edge between nodes ui and vi that takes lengthi units of time to traverse. Each node i has a disappear time given in the array disappear, indicating the time at which the node vanishes and can no longer be visited. Your task is to determine the minimum time required to reach each node from node 0 before it disappears, returning -1 if a node cannot be reached in time.

The graph may be disconnected and can contain multiple edges between the same pair of nodes. You must compute the minimum traversal times for all nodes while ensuring no node is visited after its disappearance time. Optimize your solution to handle large graphs efficiently using arrays and priority queues, taking the disappearance constraints into account.

Examples

Example 1

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

Output: [0,-1,4]

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.

Example 2

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

Output: [0,2,3]

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.

Example 3

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

Output: [0,-1]

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 Approach

Graph Construction

Build an adjacency list from the edges array. Store for each node the connected nodes along with the traversal times. Handle multiple edges by keeping the shortest traversal time for each pair to simplify Dijkstra processing.

Modified Dijkstra Traversal

Use a priority queue to explore nodes based on the earliest arrival time. Only push a node into the queue if the accumulated time is less than its disappear time. This ensures nodes that vanish cannot be incorrectly reached.

Result Compilation

Initialize the result array with -1 for all nodes. Update each node's minimum reachable time as it is popped from the priority queue. Nodes that remain -1 are unreachable due to disappearance or disconnected components.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the number of nodes and edges processed in the priority queue, generally O((n + edges.length) log n). Space complexity is dominated by the adjacency list, priority queue, and result array, typically O(n + edges.length).

What Interviewers Usually Probe

Check whether the candidate correctly handles nodes that disappear before being reached.
Listen for usage of priority queue with Dijkstra to manage earliest arrival times.
Watch if the candidate optimizes multiple edges between nodes efficiently.

Common Pitfalls or Variants

Common pitfalls

Forgetting to compare current traversal time with node disappear time, causing incorrect reachable nodes.
Not handling multiple edges between the same nodes properly, leading to suboptimal paths.
Assuming the graph is connected and not accounting for unreachable nodes.

Follow-up variants

Nodes disappear at random intervals instead of a fixed array, requiring dynamic updates during traversal.
Edges may have time-dependent weights, increasing complexity of minimum arrival computation.
Start node is not fixed at 0, requiring generalized multi-source shortest path handling.

FAQ

What is the key strategy to solve Minimum Time to Visit Disappearing Nodes efficiently?

Use a priority queue with Dijkstra's algorithm, only visiting nodes if current time is less than their disappear time.

How should multiple edges between the same nodes be handled?

Keep only the edge with the shortest traversal time to avoid suboptimal paths and unnecessary queue processing.

Can a node ever be visited after its disappear time?

No, nodes that disappear before arrival are considered unreachable and should remain -1 in the result.

Does the graph need to be connected to solve this problem?

No, disconnected nodes are allowed. Unreachable nodes will correctly remain marked as -1.

Which data structures are essential for this array plus graph problem?

An adjacency list for graph representation, an array for disappear times and results, and a priority queue for earliest arrival traversal.

terminal

Solution

Solution 1: Heap-Optimized Dijkstra

First, we create an adjacency list $\textit{g}$ to store the edges of the graph. Then, we create an array $\textit{dist}$ to store the shortest distances from node $0$ to other nodes. Initialize $\textit{dist}[0] = 0$, and the distances for the rest of the nodes are initialized to infinity.

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class Solution:
    def minimumTime(
        self, n: int, edges: List[List[int]], disappear: List[int]
    ) -> List[int]:
        g = defaultdict(list)
        for u, v, w in edges:
            g[u].append((v, w))
            g[v].append((u, w))
        dist = [inf] * n
        dist[0] = 0
        pq = [(0, 0)]
        while pq:
            du, u = heappop(pq)
            if du > dist[u]:
                continue
            for v, w in g[u]:
                if dist[v] > dist[u] + w and dist[u] + w < disappear[v]:
                    dist[v] = dist[u] + w
                    heappush(pq, (dist[v], v))
        return [a if a < b else -1 for a, b in zip(dist, disappear)]

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