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Amazon·Software Engineer·Technical Phone Screen·Intermediate

IntermediateRejected
Apr 2026

Summary

Did a technical screen for a Software Engineer role at Amazon. LP portion went fine, had my stories ready. The coding section is where things fell apart, and I'm still kind of kicking myself about it.

Questions Asked (1)

Q1

Given a network of servers where some are directly connected, find the total number of isolated clusters (groups of servers that can communicate directly or indirectly).

Algorithms & Data Structures
Author's notes

This is a connected components problem and I completely froze.

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AI HintsAI Generated

Suggested Approach

Model the servers as nodes and direct connections as edges in an undirected graph. The problem reduces to counting the number of connected components, which can be done using either BFS/DFS or Union-Find (Disjoint Set Union). Choose the method based on constraints and explain your reasoning.

Pro tip: At Amazon, interviewers value candidates who discuss trade-offs between approaches and consider edge cases like empty graphs or self-loops. Mention that Union-Find with path compression and union by rank is often more efficient for dynamic connectivity, but BFS/DFS is simpler if the graph is static.

1. Clarify the problem

Confirm that the network is undirected, connections are bidirectional, and isolated clusters are connected components. Ask about input format, constraints, and whether the graph can be modified.

2. Choose an algorithm

Decide between BFS/DFS (traverse each component) and Union-Find (merge connected nodes). Consider time/space complexity and whether the graph is static or dynamic.

3. Implement the solution

For BFS/DFS: iterate over all nodes, and for each unvisited node, perform traversal to mark its component and increment count. For Union-Find: initialize each node as its own parent, union connected nodes, then count unique roots.

4. Analyze complexity

State time and space complexity: BFS/DFS is O(V+E) time and O(V) space; Union-Find is O(E α(V)) time and O(V) space, where α is the inverse Ackermann function.

5. Test with examples

Walk through a small example, including edge cases like no edges (each node isolated) or all nodes connected (one cluster). Verify the count matches expectations.

Key Points to Mention

  • Graph representation: adjacency list or edge list
  • Connected components concept
  • BFS/DFS traversal to mark visited nodes
  • Union-Find with path compression and union by rank
  • Time and space complexity analysis
  • Edge cases: empty graph, self-loops, disconnected nodes

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.