Introduction

In this blog post, we will explore how to solve the LeetCode problem "2203" 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

You are given an integer n denoting the number of nodes of a weighted directed graph. The nodes are numbered from 0 to n - 1. You are also given a 2D integer array edges where edges[i] = [fromi, toi, weighti] denotes that there exists a directed edge from fromi to toi with weight weighti. Lastly, you are given three distinct integers src1, src2, and dest denoting three distinct nodes of the graph. Return the minimum weight of a subgraph of the graph such that it is possible to reach dest from both src1 and src2 via a set of edges of this subgraph. In case such a subgraph does not exist, return -1. A subgraph is a graph whose vertices and edges are subsets of the original graph. The weight of a subgraph is the sum of weights of its constituent edges. Example 1: Input: n = 6, edges = [[0,2,2],[0,5,6],[1,0,3],[1,4,5],[2,1,1],[2,3,3],[2,3,4],[3,4,2],[4,5,1]], src1 = 0, src2 = 1, dest = 5 Output: 9 Explanation: The above figure represents the input graph. The blue edges represent one of the subgraphs that yield the optimal answer. Note that the subgraph [[1,0,3],[0,5,6]] also yields the optimal answer. It is not possible to get a subgraph with less weight satisfying all the constraints. Example 2: Input: n = 3, edges = [[0,1,1],[2,1,1]], src1 = 0, src2 = 1, dest = 2 Output: -1 Explanation: The above figure represents the input graph. It can be seen that there does not exist any path from node 1 to node 2, hence there are no subgraphs satisfying all the constraints. Constraints: 3 <= n <= 105 0 <= edges.length <= 105 edges[i].length == 3 0 <= fromi, toi, src1, src2, dest <= n - 1 fromi != toi src1, src2, and dest are pairwise distinct. 1 <= weight[i] <= 105

Explanation

Here's the solution to the problem:

• High-Level Approach:

  1. Compute shortest distances from src1 to all nodes, from src2 to all nodes, and from all nodes to dest using Dijkstra's algorithm (forward and backward).
  2. Iterate through all nodes and find the node i that minimizes the sum of distances: dist1[i] + dist2[i] + dist_dest[i], where dist1[i] is the shortest distance from src1 to i, dist2[i] is the shortest distance from src2 to i, and dist_dest[i] is the shortest distance from i to dest.

• Complexity:

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

Code

    import heapq

def minimum_weight_subgraph(n: int, edges: list[list[int]], src1: int, src2: int, dest: int) -> int:
    """
    Finds the minimum weight of a subgraph such that it is possible to reach dest from both src1 and src2.
    """

    def dijkstra(graph, start_node):
        distances = [float('inf')] * n
        distances[start_node] = 0
        pq = [(0, start_node)]

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

            if dist > distances[u]:
                continue

            for v, weight in graph[u]:
                if distances[v] > distances[u] + weight:
                    distances[v] = distances[u] + weight
                    heapq.heappush(pq, (distances[v], v))

        return distances

    def build_graph(edges):
        graph = [[] for _ in range(n)]
        for u, v, weight in edges:
            graph[u].append((v, weight))
        return graph

    def build_reverse_graph(edges):
        graph = [[] for _ in range(n)]
        for u, v, weight in edges:
            graph[v].append((u, weight))
        return graph

    graph = build_graph(edges)
    reverse_graph = build_reverse_graph(edges)

    dist1 = dijkstra(graph, src1)
    dist2 = dijkstra(graph, src2)
    dist_dest = dijkstra(reverse_graph, dest)

    min_weight = float('inf')
    for i in range(n):
        if dist1[i] != float('inf') and dist2[i] != float('inf') and dist_dest[i] != float('inf'):
            min_weight = min(min_weight, dist1[i] + dist2[i] + dist_dest[i])

    if min_weight == float('inf'):
        return -1
    else:
        return min_weight