The bug itself was obvious once they pointed to the one-way treatment, but I fumbled a bit explaining why bidirectional matters in terms of the data structure.
First, clarify the problem by confirming that roads are bidirectional and that the adjacency list should include both directions. Then, walk through the existing code to identify where only one direction is added, and propose a fix that adds both (u -> v) and (v -> u) for each road. Finally, discuss edge cases like duplicate roads, self-loops, and disconnected components, and consider trade-offs such as using a set vs. list for neighbors.
Pro tip: Mention that you would add a test case with a simple two-node graph to verify bidirectionality, and discuss how the fix scales with large graphs. This shows attention to correctness and performance.
Confirm that roads are bidirectional and that the adjacency list should represent an undirected graph. Ask if there are any constraints like duplicate roads or self-loops.
Identify the loop that processes each road and note that it only adds an edge from one endpoint to the other. Point out the exact line where the missing reverse edge should be added.
For each road, add both directions: append the destination to the source's list and vice versa. If using a map, ensure both keys exist.
Consider duplicate roads (use a set to avoid duplicates if needed), self-loops (add twice or once depending on definition), and isolated nodes (ensure they appear in the adjacency list if required).
Write a quick test with a simple graph to verify bidirectionality. Discuss time/space complexity and any trade-offs between using lists vs. sets for neighbors.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the road network as an unweighted graph where each road is an edge of weight 1. Use BFS from the source Location to find the shortest path to the target, returning the distance or -1 if unreachable. Then state the time and space complexity as O(V + E) and O(V), respectively.
Pro tip: Clarify the graph representation (adjacency list vs. matrix) and whether the graph is directed or undirected, as this affects both the algorithm and complexity. Also, mention that BFS is optimal for unweighted graphs, but if weights were not 1, Dijkstra's algorithm would be needed.
Confirm that the graph is unweighted (all roads weight 1), and ask about graph representation, directionality, and whether the graph is connected. This ensures you understand the constraints before proposing a solution.
Explain that BFS is the ideal algorithm for finding the shortest path in an unweighted graph because it explores nodes in order of increasing distance from the source.
Describe using a queue to track nodes to visit, a visited set to avoid cycles, and a distance map or level counter to track the shortest distance from the source. Start from the source Location and stop when the target is found or the queue is exhausted.
Mention checking if source and target are the same (distance 0), if the target is unreachable (return -1), and if the graph is empty or has no edges.
State that BFS visits each vertex and edge at most once, giving O(V + E) time and O(V) space for the queue and visited set. Clarify that V is the number of locations and E is the number of roads.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the road network as a weighted graph where Location objects are nodes and road distances are edge weights. Use Dijkstra's algorithm with a priority queue to find the shortest path, handling unreachable cases by returning -1. Discuss time and space complexity, noting that non-negative weights are essential for Dijkstra's correctness.
Pro tip: Mention that if the graph is dense, a Fibonacci heap can improve Dijkstra's time complexity to O(E + V log V), but in practice a binary heap is often sufficient. Also, clarify that if the graph is unweighted or all weights are equal, BFS would be more efficient.
Confirm that the road network is a directed or undirected graph, and that distances are non-negative. Decide on an adjacency list representation for efficient traversal.
Select Dijkstra's algorithm because it handles non-negative weights and finds the shortest path from a single source. Justify why BFS or Bellman-Ford are less suitable here.
Initialize distances to infinity, set the source distance to 0, and use a min-heap to repeatedly extract the node with the smallest tentative distance. Relax outgoing edges and update distances.
After the algorithm completes, check the distance to the target. If it remains infinity, return -1; otherwise, return the distance.
State that with a binary heap, time complexity is O((V + E) log V) and space is O(V + E). Mention alternative implementations (e.g., Fibonacci heap) and their trade-offs.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.