Design Graph With Shortest Path Calculator

Hard

Samsung

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Topics:

GraphsGreedy Algorithms

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] = [from_i, to_i, edgeCost_i] meaning that there is an edge from from_i to to_i with the cost edgeCost_i.

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 <= from_i, to_i, from, to, node1, node2 <= n - 1
1 <= edgeCost_i, 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.

Design Graph With Shortest Path Calculator

Hard

Samsung

0 views

Topics:

GraphsGreedy Algorithms

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] = [from_i, to_i, edgeCost_i] meaning that there is an edge from from_i to to_i with the cost edgeCost_i.

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 <= from_i, to_i, from, to, node1, node2 <= n - 1
1 <= edgeCost_i, 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

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the number of nodes `n` and the number of edges in the input `edges`? What is the maximum cost of an edge?
Are the node IDs 0-indexed or 1-indexed? Also, can there be multiple edges between the same pair of nodes, and if so, how should I handle them (e.g., take the minimum cost, sum the costs, etc.)?
Can the graph be disconnected? If a path does not exist between the start and end nodes in `shortestPath`, is returning -1 the only acceptable output, or are there other error codes or exceptions I should consider?
In the `addEdge` method, should I assume that the nodes involved are always valid (i.e., within the range [0, n-1] or [1, n] depending on the indexing) or do I need to handle cases where invalid node IDs are provided?
Does the cost of an edge need to be non-negative? Is Dijkstra's the only acceptable algorithm or can I consider other shortest path algorithms if applicable?

Brute Force Solution

Approach

To find the shortest path in a graph using brute force, we explore every possible route between the starting point and the destination. We calculate the length of each route and then choose the shortest one. This method guarantees finding the shortest path if it exists, but can take a very long time for larger graphs.

Here's how the algorithm would work step-by-step:

List every possible path from the starting point to the destination.
Calculate the total distance of each path by adding up the distances between each connected point on the path.
Compare all the calculated distances.
Identify the path with the smallest total distance.
That path with the smallest distance is the shortest path.

Code Implementation

def find_shortest_path_brute_force(graph, start_node, end_node):
    all_possible_paths = list_all_paths(graph, start_node, end_node)

    shortest_path = None
    shortest_distance = float('inf')

    for path in all_possible_paths:
        current_path_distance = calculate_path_distance(graph, path)

        # Check if current path is the shortest
        if current_path_distance < shortest_distance:

            shortest_distance = current_path_distance
            shortest_path = path

    return shortest_path

def list_all_paths(graph, start_node, end_node, path=None):
    if path is None:
        path = [start_node]

    if start_node == end_node:
        return [path]

    if start_node not in graph:
        return []

    paths = []
    for neighbor in graph[start_node]:
        if neighbor not in path:

            new_path = path + [neighbor]
            new_paths = list_all_paths(graph, neighbor, end_node, new_path)
            paths.extend(new_paths)
    return paths

def calculate_path_distance(graph, path):
    total_distance = 0
    for i in range(len(path) - 1):
        current_node = path[i]
        next_node = path[i + 1]

        # Ensure edge exists to avoid errors
        if next_node not in graph[current_node]:

            return float('inf')

        total_distance += graph[current_node][next_node]
    return total_distance

#Example graph representation
#graph = {
#    'A': {'B': 5, 'C': 2},
#    'B': {'D': 4},
#    'C': {'B': 8, 'D': 7},
#    'D': {}
#}
#
#start_node = 'A'
#end_node = 'D'
#
#shortest_path = find_shortest_path_brute_force(graph, start_node, end_node)
#
#if shortest_path:
#    print("Shortest path:", shortest_path)
#else:
#    print("No path found.")

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(V!) – The brute force approach explores every possible path between the starting node and the destination node. In the worst-case scenario, the algorithm may need to enumerate all possible permutations of the vertices (V) in the graph to find the shortest path. Generating all permutations of V vertices takes O(V!) time. Calculating the length of each path takes O(V) time in the worst case, but this does not change the overall complexity since V! dominates V. Therefore, the overall time complexity is O(V!).

Space Complexity

O(N!) – The brute force approach, as described, explores every possible path. To list every possible path from the starting point to the destination, the algorithm implicitly stores these paths, potentially using data structures like lists or arrays. In the worst case, if the graph is dense, the number of paths can grow factorially with the number of nodes, N. Therefore, the auxiliary space required to store all possible paths grows as O(N!).

Optimal Solution

Approach

To efficiently design a graph with shortest path calculations, we'll use a well-known algorithm optimized for finding the quickest route between any two points. We'll start by creating the graph and then leverage this algorithm to answer shortest path queries efficiently, avoiding recalculating paths every time.

Here's how the algorithm would work step-by-step:

First, store the graph's structure using a method that makes it easy to access neighboring nodes and the distances between them.
Next, when the graph is created, immediately calculate the shortest paths between all pairs of points in the graph. This one-time calculation will significantly speed up subsequent queries.
Utilize an efficient shortest-path finding algorithm to precompute all possible shortest paths. This algorithm explores the graph in a smart way, always choosing the path that seems shortest so far and updating its estimates as it goes.
Store these precomputed shortest paths in a lookup table, which allows for quick retrieval of the shortest path length between any two points without needing to recompute the path each time a query is made.
When someone asks for the shortest path between two points, simply look up the precomputed value in your stored lookup table and return it. This offers almost instant results, as the hard work has already been completed.

Code Implementation

import heapq

class Graph:
    def __init__(self, number_of_nodes):
        self.number_of_nodes = number_of_nodes
        self.adjacency_list = [[] for _ in range(number_of_nodes)]

    def add_edge(self, node1, node2, weight):
        self.adjacency_list[node1].append((node2, weight))

class GraphWithShortestPaths:
    def __init__(self, number_of_nodes):
        self.graph = Graph(number_of_nodes)
        self.shortest_paths = {}

    def add_edge(self, node1, node2, weight):
        self.graph.add_edge(node1, node2, weight)

    def precompute_shortest_paths(self):
        # Precompute shortest paths between all pairs of nodes.
        for start_node in range(self.graph.number_of_nodes):
            self.shortest_paths[start_node] = self.dijkstra(start_node)

    def dijkstra(self, start_node):
        # Implement Dijkstra's algorithm for single-source shortest paths.
        distances = {node: float('inf') for node in range(self.graph.number_of_nodes)}
        distances[start_node] = 0
        priority_queue = [(0, start_node)]

        while priority_queue:
            distance, current_node = heapq.heappop(priority_queue)

            if distance > distances[current_node]:
                continue

            for neighbor, weight in self.graph.adjacency_list[current_node]:
                new_distance = distance + weight

                # Update distance if a shorter path is found.
                if new_distance < distances[neighbor]:
                    distances[neighbor] = new_distance
                    heapq.heappush(priority_queue, (new_distance, neighbor))

        return distances

    def get_shortest_path(self, node1, node2):
        # Retrieve precomputed shortest path length.
        if not self.shortest_paths:
            self.precompute_shortest_paths()

        # Return infinity if there is no path
        if self.shortest_paths[node1][node2] == float('inf'):
            return -1

        return self.shortest_paths[node1][node2]

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n^3) – The graph's structure is stored, allowing efficient neighbor access. Floyd-Warshall algorithm is used to precompute all shortest paths. The algorithm iterates through all possible intermediate nodes (n), and for each intermediate node, it considers all possible start and end nodes (n*n), resulting in nested loops of size n each. Thus, the overall time complexity is O(n*n*n), which simplifies to O(n^3).

Space Complexity

O(N^2) – The primary auxiliary space usage comes from storing the precomputed shortest paths between all pairs of nodes. This is mentioned as storing the shortest paths in a 'lookup table'. Assuming there are N nodes in the graph, this lookup table will be a matrix of size N x N to store the shortest path length between each pair of nodes. Therefore, the space required to store this matrix is proportional to N^2, leading to a space complexity of O(N^2).

Edge Cases

Case	How to Handle
Null or empty edges list in constructor.	Initialize the graph with no edges if the input edges list is null or empty.
n is zero or negative.	Raise an IllegalArgumentException since the number of nodes should be a positive integer.
Edges contain negative costs.	Raise an IllegalArgumentException as Dijkstra's algorithm requires non-negative edge weights.
addEdge called with negative cost.	Raise an IllegalArgumentException since negative costs are not allowed.
addEdge called with nodes outside the range [0, n-1].	Raise an IllegalArgumentException because node indices must be within the valid range.
shortestPath called with start or end nodes outside the range [0, n-1].	Return -1 since the start and end nodes must be valid within the graph.
No path exists between start and end nodes in shortestPath.	Dijkstra's algorithm should naturally return -1 if the end node is never reached, indicating no path exists.
Graph contains cycles.	Dijkstra's algorithm handles cycles correctly by always choosing the shortest known path, effectively ignoring longer cyclical paths.