2642 - Design Graph With Shortest Path Calculator\.

Hard

There is a directed weighted graph that consists of n nodes numbered from 0 to n - 1. The edges of the graph are initially represented by the given array edges where <code>edgesi = from_i, to_i, edgeCost_i</code> meaning that there is an edge from <code>from_i</code> to <code>to_i</code> with the cost <code>edgeCost_i</code>.

Implement the Graph class:

• Graph(int n, int[][] edges) initializes the object with n nodes and the given edges.

• addEdge(int[] edge) adds an edge to the list of edges where edge = [from, to, edgeCost]. It is guaranteed that there is no edge between the two nodes before adding this one.

• int shortestPath(int node1, int node2) returns the minimum cost of a path from node1 to node2. If no path exists, return -1. The cost of a path is the sum of the costs of the edges in the path.

Example 1:

Input "Graph", "shortestPath", "shortestPath", "addEdge", "shortestPath" [[4, [0, 2, 5, 0, 1, 2, 1, 2, 1, 3, 0, 3]], 3, 2, 0, 3, [1, 3, 4], 0, 3]

Output: null, 6, -1, null, 6

Explanation:

Graph g = new Graph(4, [0, 2, 5, 0, 1, 2, 1, 2, 1, 3, 0, 3]);

g.shortestPath(3, 2); // return 6. The shortest path from 3 to 2 in the first diagram above is 3 -> 0 -> 1 -> 2 with a total cost of 3 + 2 + 1 = 6.

g.shortestPath(0, 3); // return -1. There is no path from 0 to 3.

g.addEdge(1, 3, 4); // We add an edge from node 1 to node 3, and we get the second diagram above.

g.shortestPath(0, 3); // return 6. The shortest path from 0 to 3 now is 0 -> 1 -> 3 with a total cost of 2 + 4 = 6.

Constraints:

• 1 <= n <= 100

• 0 <= edges.length <= n * (n - 1)

• edges[i].length == edge.length == 3

• <code>0 <= from_i, to_i, from, to, node1, node2 <= n - 1</code>

• <code>1 <= edgeCost_i, edgeCost <= 10⁶</code>

• There are no repeated edges and no self-loops in the graph at any point.

• At most 100 calls will be made for addEdge.

• At most 100 calls will be made for shortestPath.