3112. Minimum Time to Visit Disappearing Nodes

Problem Link

Description

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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.

Note 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

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Python3

class Solution:
    def minimumTime(self, n: int, edges: List[List[int]], disappear: List[int]) -> List[int]:
        dist = [inf] * n
        graph = defaultdict(list)
        
        for x, y, length in edges:
            graph[x].append((y, length))
            graph[y].append((x, length))
 
        pq = [(0, 0)]
        dist[0] = 0
        while pq:
            d, node = heappop(pq)
 
            if d != dist[node]: continue
 
            for adj, d2 in graph[node]:
                new = d + d2
                if new < dist[adj] and new < disappear[adj]:
                    dist[adj] = new
                    heappush(pq, (new, adj))
        
        for i in range(n):
            if dist[i] == inf:
                dist[i] = -1
 
        return dist