How do I practice coding and algorithm questions?

Use PracHub's coding console to write, test, and debug your solutions in Python or JavaScript. View hints, test against sample inputs, and compare with official solutions.

What difficulty level is this coding question?

This is a easy difficulty Coding & Algorithms question, commonly asked during Technical Screen rounds at Bloomberg.

What role is this question designed for?

This question is commonly asked for Software Engineer candidates at Bloomberg during technical interviews.

Check connectivity between two subway stations

Quick Overview

This problem evaluates graph connectivity and traversal concepts—specifically reasoning about adjacency representations, path existence, cycles, disconnected components, and missing-node edge cases—within the Coding & Algorithms domain at an implementation-level (medium) abstraction.

Company: Bloomberg

Role: Software Engineer

Category: Coding & Algorithms

Difficulty: easy

Interview Round: Technical Screen

You are given a description of subway stations A, B, C, D, E, ... as a graph. For each station, you are told which other stations it directly connects to (e.g., an adjacency list/map). Write a function that, given two station names start and end, returns whether it is possible to travel from start to end by following one or more connections (i.e., whether a path exists). Assume station names are strings. If a station is not present in the map, treat it as having no outgoing connections. Additionally, write a few test cases for your function (including cases with cycles, disconnected components, start==end, and missing stations).

Quick Answer: This problem evaluates graph connectivity and traversal concepts—specifically reasoning about adjacency representations, path existence, cycles, disconnected components, and missing-node edge cases—within the Coding & Algorithms domain at an implementation-level (medium) abstraction.

You are given a subway network as an adjacency map where each key is a station name and its value is a list of stations that can be reached directly from it. Write a function that returns whether it is possible to travel from a given start station to a given end station by following these connections. Treat the graph exactly as listed: you may only travel from a station to the stations in its adjacency list. If a connection is bidirectional, both directions must appear in the map. If a station is missing from the map, treat it as having no outgoing connections. A station is considered reachable from itself, even without taking any connections. The graph may contain cycles.

Constraints

0 <= number of stations in the map <= 100000
0 <= total number of listed connections <= 200000
Station names are strings
A station may appear in a neighbor list even if it is not a key in the map; such a station has no outgoing connections

Examples

Input: ({'A': ['B', 'C'], 'B': ['D'], 'C': [], 'D': ['E'], 'E': []}, 'A', 'E')

Expected Output: True

Explanation: One valid path is A -> B -> D -> E.

Input: ({'A': ['B'], 'B': ['C'], 'C': ['A', 'D'], 'D': []}, 'A', 'D')

Expected Output: True

Explanation: The graph contains a cycle A -> B -> C -> A, but D is still reachable through A -> B -> C -> D.

Input: ({'A': ['B'], 'B': [], 'C': ['D'], 'D': []}, 'A', 'D')

Expected Output: False

Explanation: A and D are in different disconnected components.

Input: ({'A': ['B'], 'B': ['A']}, 'B', 'B')

Expected Output: True

Explanation: A station is always reachable from itself.

Input: ({'A': ['B'], 'B': []}, 'X', 'B')

Expected Output: False

Explanation: X is not in the map, so it has no outgoing connections and cannot reach B.

Input: ({'A': ['B'], 'B': ['X']}, 'A', 'X')

Expected Output: True

Explanation: X is not a key in the map, but it can still be reached because B connects directly to X.

Input: ({}, 'A', 'B')

Expected Output: False

Explanation: With an empty graph, there are no connections to follow.

Hints

This is a graph reachability problem. Try exploring from the start station using DFS or BFS.
Keep a visited set so you do not get stuck in cycles by revisiting the same station repeatedly.

Quick Overview