Uber Interview Review: Topological Sort Traps?

🔥 Recent Uber SDE Interview Record

In a recent Uber technical interview, our client encountered a classic but "trap-filled" problem — Package Build Order.

At first glance, it looks like a simple graph traversal, but it requires a solid understanding of Topological Sort. Many candidates fall into the BFS Level Traversal Trap, failing to handle deep dependency orders correctly.

With oavoservice's real-time assistance, the client corrected their initial wrong approach and wrote a clear, logically perfect solution, earning high praise from the interviewer.

📄 Problem Description

Given a package name, return the build order of its package dependencies. If package A depends on package B (A -> B), then B must be built before A.

Example Logic:

• Target: Service
• Dependencies: Service -> Core, Core -> Types, Types -> Interfaces ...
• Expected Output: ["Interfaces", "Types", "Adapters", "Core", "Utils", "Service"]

🛑 The "BFS Layer" Trap

Many candidates think: "I just need to find all dependencies, layer by layer."

Wrong Approach (BFS):

1. Find dependencies of Service -> [Core, Utils]
2. Build Core, Utils
3. Find dependencies of Core -> [Types]...

Fatal Flaw: If Utils also depends on Types (Utils -> Types), BFS might build Utils before Types is ready! This violates the build rule.

Correct Model: This is a Topological Sort problem on a Directed Acyclic Graph (DAG).

✅ oavoservice Real-time Guidance

Phase 1: Clarification & Modeling (Crucial!)

oavoservice Hint:

"Don't use BFS layers. This is a dependency resolution problem. You need Topological Sort." "Confirm with the interviewer: Can there be circular dependencies?" (If yes, handle cycle detection).

Phase 2: Implementation (DFS Post-order)

We guided the client to use DFS Post-order Traversal because it naturally fits the "Build dependencies first, then build self" logic.

Core Logic:

1. To build A, first visit all children of A.
2. After visiting all children, add A to the result list.
3. Use a visited set to avoid reprocessing and detect cycles.

Code Snippet:

def get_build_order(package, dependencies):
    visited = set()
    building = set()  # Detect cycles
    order = []

    def dfs(node):
        if node in building:
            raise ValueError("Cycle detected!")
        if node in visited:
            return
        
        building.add(node)
        for dep in dependencies.get(node, []):
            dfs(dep)
        
        building.remove(node)
        visited.add(node)
        order.append(node)

    dfs(package)
    return order

Phase 3: Follow-up & Optimization

Interviewer: "What if we want to build in parallel?"

oavoservice Hint:

"Mention Kahn's Algorithm (in-degree based). Nodes with in-degree 0 can be built in parallel."

🎓 Why Choose oavoservice?

• Real-time Course Correction: Stop you from going down the wrong path (like using BFS for TopoSort).
• Professional Communication: Teach you how to articulate "Dependency Graph", "DAG", "Cycle Detection".
• Complete Solution: Not just code, but handling edge cases and scalability.

Whether it's Uber, Amazon, Meta, or TikTok, oavoservice is your reliable technical backup.

Need Interview Assistance? Contact Us

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