2642. Design Graph With Shortest Path Calculator

Problem Link

Description

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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 edges[i] = [fromi, toi, edgeCosti] meaning that there is an edge from fromi to toi with the cost edgeCosti.

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
• 0 <= fromi, toi, from, to, node1, node2 <= n - 1
• 1 <= edgeCosti, edgeCost <= 106
• 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.

Solution

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Python3

class Graph:
 
    def __init__(self, n: int, edges: List[List[int]]):
        self.N = n
        self.graph = defaultdict(list)
 
        for a, b, cost in edges:
            self.graph[a].append((b, cost))
 
    def addEdge(self, edge: List[int]) -> None:
        a, b, cost = edge
        self.graph[a].append((b, cost))
 
    def shortestPath(self, node1: int, node2: int) -> int:
        dist = [inf] * self.N
        dist[node1] = 0
        pq = [(0, node1)]
 
        while pq:
            weight, node = heappop(pq)
 
            if dist[node] != weight: continue
 
            for adj, w in self.graph[node]:
                if weight + w < dist[adj]:
                    dist[adj] = weight + w
                    heappush(pq, (dist[adj], adj))
 
        if dist[node2] == inf: return -1
 
        return dist[node2]
 
 
# Your Graph object will be instantiated and called as such:
# obj = Graph(n, edges)
# obj.addEdge(edge)
# param_2 = obj.shortestPath(node1,node2)