Problem Description

In this problem, we are given a directed weighted graph represented as a 2D array called edges, with each sub-array representing an edge in the format [start_node, end_node, weight]. The graph has n nodes that are 0-indexed, which means that nodes are numbered from 0 to n-1. We are also given a starting node s and an array of nodes called marked.

The objective is to determine the minimum distance from the starting node s to any node within the array marked. To clarify, the 'distance' here refers to the sum of the weights along the shortest path from s to the target node.

If there's no possible path from s to any of the nodes in marked, we return -1. The challenge lies in computing the shortest paths in an efficient manner, taking into consideration that the graph may contain edge cases like loops or multiple edges between two nodes.

Intuition

For the solution, we need an approach that allows us to find the shortest path from a source node to all other nodes in a graph. This points us towards the Dijkstra's algorithm, which is specifically designed for this purpose. However, we modify it slightly to use an adjacency matrix rather than a typical adjacency list, which is more efficient when dealing with dense graphs.

Here's the intuition behind the solution:

1. We initially set a graph matrix g with dimensions nxn, where g[u][v] will hold the minimum weight of all edges from node u to node v. If there are multiple edges between two nodes, we only keep the one with the least weight.

2. A distance array dist is created to keep track of the shortest distance from the start node s to every other node. Initially, all distances are set to infinity (inf), except for the distance from s to itself, which is 0.

3. We also have a boolean array vis to mark nodes as visited once their shortest distance is finalized.

4. The algorithm repeatedly finds the node with the smallest known distance that hasn't been visited yet. This node is marked as visited, and we examine all of its neighbors.

5. For each neighbor, we update its distance in the dist array if we find a shorter path through the current node. This step is repeated until distances to all nodes are finalized.

6. Finally, we find the minimum distance to all marked nodes and return that value, unless the minimum distance is infinity, which indicates that the nodes in marked are not reachable from s. In such a case, we return -1.

By using this modified version of Dijkstra's algorithm, we ensure that we check all possible paths in an efficient manner and determine the shortest path from s to a node in marked.

Learn more about Graph, Shortest Path and Heap (Priority Queue) patterns.

Solution Approach

The solution uses a modified Dijkstra's algorithm to find the shortest path from a single source to all nodes and then specifically to a subset of those nodes (marked nodes). Let's walk through the key steps of the implementation based on the provided python code:

1. Initialize the Graph Matrix: A graph matrix g is created to represent the weighted edges of the graph. It's initialized with inf (representing an absence of direct connection between nodes) and filled with actual weights where edges are present.

   g = [[inf] * n for _ in range(n)]
   for u, v, w in edges:
       g[u][v] = min(g[u][v], w)

2. Set Initial Distances: dist array is set with inf values indicating unknown shortest paths. The distance from the source node s to itself is set to 0, as the shortest path from a node to itself is by definition no movement.

   dist = [inf] * n
   dist[s] = 0

3. Node Visitation Tracking: A boolean array vis keeps track of which nodes' shortest paths have been solidified. Initially, all nodes are set to False to indicate that no node's path has been finalized.

   vis = [False] * n

4. Algorithm Loop: The code executes a loop that iterates n times. Each iteration selects the non-visited node with the smallest known distance from the dist array.

   for _ in range(n):
       t = -1
       for j in range(n):
           if not vis[j] and (t == -1 or dist[t] > dist[j]):
               t = j

5. Update Shortest Paths: For the selected node, the algorithm iterates over all nodes and updates their shortest paths if a shorter path is found through the selected node.

   vis[t] = True
   for j in range(n):
       dist[j] = min(dist[j], dist[t] + g[t][j])

6. Find Minimum Distances to Marked Nodes: After the loop ends, the distances to all marked nodes are found. It selects the smallest value from the dist array among the indices that correspond to the marked nodes.

   ans = min(dist[i] for i in marked)

7. Return the Result: The algorithm returns the shortest distance unless it is inf (in which case, it returns -1) because inf indicates no path was found from source s to any of the marked nodes.

   return -1 if ans >= inf else ans

The key data structures used here include a 2D list g for the graph representation, a list dist for tracking the shortest path distances, and vis, a list for tracking node visitation. The pattern used is iterative selection and relaxation of distances, characteristic of Dijkstra's algorithm.

Example Walkthrough

Let's consider a small graph with n = 4 nodes and s = 0 as the starting node. We have the edges and weights defined as follows:

• edges = [[0, 1, 2], [0, 2, 5], [1, 2, 1], [2, 3, 1]]
• marked = [2, 3]

The goal is to find the minimum distance from node 0 to any node in marked. Following the solution approach:

1. Initialize the Graph Matrix: We initialize a graph matrix g of size 4x4 with inf, except where there are actual connections:

   g = [
       [inf,   2,   5, inf],
       [inf, inf,   1, inf],
       [inf, inf, inf,   1],
       [inf, inf, inf, inf]
   ]

   We only keep the smallest weights between nodes.

2. Set Initial Distances: The dist array is initialized with inf, but we set dist[0] to 0 because that is our start node:

   dist = [0, inf, inf, inf]

3. Node Visitation Tracking: We track which nodes have been visited in an array vis, with all elements initially set to False:

   vis = [False, False, False, False]

4. Algorithm Loop: We loop through the nodes to find the non-visited node with the smallest distance each time. As we start with s = 0, it is the first to be selected and marked visited:

   vis = [True, False, False, False]  # Mark node 0 as visited

5. Update Shortest Paths: We update the shortest path to all the other nodes from node 0. After the first iteration, due to directly connected edges, dist array would be updated as follows:

   dist = [0, 2, 5, inf]

   Continuing with the algorithm, node 1 would be chosen next and dist would be updated to:

   dist = [0, 2, 3, inf]

   Then, node 2 would be chosen, and dist would be updated to:

   dist = [0, 2, 3, 4]

6. Find Minimum Distances to Marked Nodes: After processing all nodes, we look at the dist values of the marked nodes [2, 3], which are 3 and 4, respectively.

7. Return the Result: We select the minimum distance to a marked node, which is 3 for node 2, and since it's not inf, we return 3:

   return 3

In this small example, the minimum distance from node 0 to any node in marked is 3, which is the distance from node 0 to node 2.

Solution Implementation

Time and Space Complexity

Time Complexity

The given code implements a modified version of the Dijkstra's algorithm for finding the shortest path from a single source to all other vertices in a graph.

• Initializing the graph 'g', 'dist', and 'vis' lists takes O(n^2), O(n), and O(n) time complexities respectively, where 'n' is the number of nodes in the graph.

• The outer for loop runs for 'n' iterations, and within this loop:

  • Selecting the node 't' with the minimum distance 'dist[t]' that is not visited yet, requires iterating over all 'n' nodes. So it takes O(n) time per iteration, resulting in O(n^2) for all 'n' iterations.
  • The nested inner for loop also iterates over 'n' nodes to update the distance 'dist[j]'. Therefore, it takes O(n) time per outer loop iteration, resulting in O(n^2) for all 'n' iterations of the outer loop.

Combining these together, the time complexity of the code is dominated by the two nested 'n' iterations, resulting in a total time complexity of O(n^2 + n^2), which simplifies to O(n^2).

Space Complexity

• The space complexity for the graph representation 'g' is O(n^2) since it stores the weights for each pair of vertices.
• 'dist' and 'vis' arrays each take O(n) space.

Thus, the overall space complexity of the code is the sum of the space taken by 'g', 'dist', and 'vis', resulting in O(n^2 + n + n), which simplifies to O(n^2).

Learn more about how to find time and space complexity quickly using problem constraints.