LeetCode in Kotlin

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 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

import java.util.PriorityQueue

class Graph(n: Int, edges: Array<IntArray>) {
    private val adj = HashMap<Int, ArrayList<Pair<Int, Int>>>().apply {
        for (i in 0 until n)
            this[i] = ArrayList<Pair<Int, Int>>()

        for ((u, v, cost) in edges) {
            this[u] = getOrDefault(u, ArrayList<Pair<Int, Int>>()).apply { this.add(v to cost) }
        }
    }

    fun addEdge(edge: IntArray) {
        val (u, v, cost) = edge
        adj[u] = adj.getOrDefault(u, ArrayList<Pair<Int, Int>>()).apply { this.add(v to cost) }
    }

    fun shortestPath(node1: Int, node2: Int): Int {
        val minHeap = PriorityQueue<Pair<Int, Int>> { a, b -> a.second - b.second }
        val distance = IntArray(adj.size) { Integer.MAX_VALUE }
        minHeap.add(node1 to 0)
        distance[node1] = 0
        while (minHeap.isNotEmpty()) {
            val (node, cost) = minHeap.poll()
            if (node == node2) return cost
            if (cost > distance[node]) continue
            adj[node]?.let {
                for ((next, nextCost) in adj[node]!!) {
                    if (cost + nextCost < distance[next]) {
                        distance[next] = cost + nextCost
                        minHeap.add(next to cost + nextCost)
                    }
                }
            }
        }
        return -1
    }
}

/*
 * Your Graph object will be instantiated and called as such:
 * var obj = Graph(n, edges)
 * obj.addEdge(edge)
 * var param_2 = obj.shortestPath(node1,node2)
 */

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