Introduction

In this blog post, we will explore how to solve the LeetCode problem "3112" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the breakdown of the solution:

• Dijkstra's Algorithm: Use Dijkstra's algorithm to find the shortest path from node 0 to all other nodes.
• Disappearance Time Constraint: During the traversal, ensure that the time taken to reach a node is strictly less than its disappearance time. If the time is equal to or greater than the disappearance time, the node is unreachable.

• Graph Representation: Represent the graph using an adjacency list where each node stores a list of its neighbors along with the edge length.

• Runtime Complexity: O(E log V), where E is the number of edges and V is the number of vertices (nodes). Storage Complexity: O(V + E).

Code

    import heapq

def solve():
    n = int(input())
    edges_str = input()
    disappear_str = input()

    edges = eval(edges_str)
    disappear = eval(disappear_str)

    adj = [[] for _ in range(n)]
    for u, v, length in edges:
        adj[u].append((v, length))
        adj[v].append((u, length))

    dist = [-1] * n
    dist[0] = 0

    pq = [(0, 0)]  # (distance, node)

    while pq:
        d, u = heapq.heappop(pq)

        if d > dist[u] and dist[u] != -1:
            continue

        if d >= disappear[u]:
            continue

        for v, weight in adj[u]:
            if d + weight < disappear[v]:
                if dist[v] == -1 or dist[v] > d + weight:
                    dist[v] = d + weight
                    heapq.heappush(pq, (dist[v], v))

    print(dist)

# Example Usage (You can uncomment these lines and adjust the input)
# n = 3
# edges = [[0, 1, 2], [1, 2, 1], [0, 2, 4]]
# disappear = [1, 1, 5]
# Input:
# 3
# [[0,1,2],[1,2,1],[0,2,4]]
# [1,1,5]

# Input:
# 3
# [[0,1,2],[1,2,1],[0,2,4]]
# [1,3,5]

# Input:
# 2
# [[0,1,1]]
# [1,1]

solve()