How do you solve Minimum Time to Visit Disappearing Nodes on LeetCode?

Minimum Time to Visit Disappearing Nodes (LeetCode #3112) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(E log V) time complexity and O(V) space complexity. Key insight: Use Dijkstra's algorithm for shortest paths with constraints.

What is the brute force solution for Minimum Time to Visit Disappearing Nodes?

The brute force approach for Minimum Time to Visit Disappearing Nodes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible paths from the starting node to each node and calculating the time taken. If the time taken to reach a node exceeds its disappearance time, we cannot visit it. The optimal Optimal Solution approach improves this to O(E log V).

What algorithm or data structure is used to solve Minimum Time to Visit Disappearing Nodes?

Minimum Time to Visit Disappearing Nodes uses the following concepts: Array, Graph Theory, Heap (Priority Queue), Shortest Path. The recommended patterns to study are: Graph Traversal, Priority Queue, Dijkstra's Algorithm.

What are the interview tips for Minimum Time to Visit Disappearing Nodes?

Explain your thought process clearly while coding. Discuss edge cases, such as disconnected graphs. Practice implementing Dijkstra's algorithm.

What are common mistakes when solving Minimum Time to Visit Disappearing Nodes?

Forgetting to check if a node can be visited before it disappears. Not using a priority queue effectively for pathfinding.

#3112

Minimum Time to Visit Disappearing Nodes

Medium

ArrayGraph TheoryHeap (Priority Queue)Shortest PathGraph TraversalPriority QueueDijkstra's Algorithm

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(E log V)
Space	O(1)	O(V)

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Intuition

Time O(E log V)Space O(V)

We can use Dijkstra's algorithm to find the shortest path to each node while considering the disappearance times. This approach efficiently finds the minimum time to reach each node by prioritizing paths based on their traversal time.

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Algorithm

3 steps

1Step 1: Initialize a priority queue and a result array with -1 for all nodes except the starting node (node 0) which is set to 0.
2Step 2: Push the starting node into the priority queue with a time of 0.
3Step 3: While the priority queue is not empty, pop the node with the smallest time. If this time is less than the disappearance time, update the result for that node and push its neighbors into the queue with updated times.

solution.py26 lines

1# Full working Python code
2import heapq
3from collections import defaultdict
4
5def min_time_to_visit(n, edges, disappear):
6    graph = defaultdict(list)
7    for u, v, length in edges:
8        graph[u].append((v, length))
9        graph[v].append((u, length))
10
11    result = [-1] * n
12    result[0] = 0
13    pq = [(0, 0)]  # (time, node)
14
15    while pq:
16        time, node = heapq.heappop(pq)
17        if time >= disappear[node]:
18            continue
19        for neighbor, travel_time in graph[node]:
20            new_time = time + travel_time
21            if new_time < disappear[neighbor] and (result[neighbor] == -1 or new_time < result[neighbor]):
22                result[neighbor] = new_time
23                heapq.heappush(pq, (new_time, neighbor))
24
25    return result
26

ℹ

Complexity note: The time complexity is O(E log V) because we use a priority queue to efficiently get the next node with the smallest time, where E is the number of edges and V is the number of nodes.

1Use Dijkstra's algorithm for shortest paths with constraints.
2Consider node disappearance times when updating paths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.