Google·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Google coding interview, one question about spiral matrix traversal. Pretty standard algorithms session, nothing too surprising.

Questions Asked (1)

Given an m x n matrix, write a function to return all elements in spiral order.

Algorithms & Data Structures

Author's notes

Classic problem but I still fumbled the boundary conditions a bit.

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AI HintsAI Generated

Suggested Approach

Clarify the problem and edge cases, then propose a layer-by-layer traversal using four boundaries (top, bottom, left, right). Walk through the algorithm with a small example, analyze time and space complexity, and discuss potential optimizations or alternative approaches.

Pro tip: Mention that you would test with edge cases like empty matrix, single row/column, and non-square matrices, and discuss how to handle them gracefully. This shows attention to detail and robustness.

1. Clarify and Confirm

Ask clarifying questions: Is the matrix guaranteed non-empty? What are the constraints on m and n? Should the output be a list or array? Confirm the expected order (clockwise spiral starting from top-left).

2. Outline Approach

Propose a boundary-based simulation: maintain four pointers (top, bottom, left, right) and traverse the matrix in spiral order by moving along the top row, right column, bottom row, and left column, then shrinking the boundaries.

3. Walk Through Example

Trace the algorithm on a small example (e.g., 3x3 matrix) to demonstrate correctness and show how boundaries are updated after each direction.

4. Analyze Complexity

State that time complexity is O(m*n) since each element is visited once, and space complexity is O(1) extra space (excluding the output array).

5. Discuss Edge Cases and Testing

Mention edge cases: empty matrix, single row, single column, and how the algorithm handles them. Suggest writing unit tests for these cases.

Key Points to Mention

Boundary variables: top, bottom, left, right initialized to 0, m-1, 0, n-1 respectively.
Traversal order: left-to-right along top row, top-to-bottom along right column, right-to-left along bottom row (if top <= bottom), bottom-to-top along left column (if left <= right).
Update boundaries after each direction: top++, right--, bottom--, left++.
Termination condition: continue while top <= bottom and left <= right.
Time complexity O(m*n), space complexity O(1) extra space.
Handling edge cases: empty matrix returns empty list; single row/column handled by boundary checks.

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