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Xiaopeng·Software Engineer·Technical Phone Screen·Intermediate

IntermediateRejected
Apr 2026Remote

Summary

Technical phone screen for a software engineer role at Xiaopeng. No recruiter call beforehand, jumped straight into a hard coding problem. The interviewer was pleasant but the rejection email came two days later anyway.

Questions Asked (1)

Q1

Given a matrix, find the maximum sum rectangle no larger than a target value (LeetCode Hard 363).

Algorithms & Data Structures
Author's notes

My first instinct was slightly wrong.

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AI HintsAI Generated

Suggested Approach

Reduce the 2D problem to 1D by fixing left and right column boundaries and compressing each row into a sum, then find the maximum subarray sum no larger than the target using a sorted list (e.g., TreeSet) for efficient prefix sum lookups. This yields O(min(m,n)^2 * max(m,n) * log(max(m,n))) time, which is optimal for this problem.

Pro tip: Mention that you can transpose the matrix to ensure the smaller dimension is squared, reducing time complexity when the matrix is highly rectangular. Also, clarify that the sorted list approach is necessary because the target constraint prevents using Kadane's algorithm directly.

1. Clarify constraints and edge cases

Ask about matrix dimensions, value ranges, and whether the target can be negative. Discuss handling of empty matrix or no valid rectangle.

2. Reduce to 1D subproblem

Explain that by fixing left and right columns, each row's sum between those columns can be computed, turning the problem into finding the maximum subarray sum ≤ target in a 1D array.

3. Solve 1D with prefix sums and sorted list

Use a sorted list (e.g., TreeSet) to store prefix sums. For each prefix sum, find the smallest prefix sum ≥ current - target to maximize the subarray sum without exceeding target.

4. Optimize with transpose

If the number of rows is greater than columns, transpose the matrix to minimize the squared dimension, improving time complexity.

5. Analyze complexity and test

State time complexity O(min(m,n)^2 * max(m,n) * log(max(m,n))) and space O(max(m,n)). Walk through a small example to verify correctness.

Key Points to Mention

  • Reduction from 2D to 1D by fixing column boundaries and compressing rows.
  • Use of prefix sums to compute subarray sums in O(1).
  • Sorted list (TreeSet) for efficient ceiling queries to enforce the target constraint.
  • Handling of negative numbers and the need for a sorted structure (Kadane's algorithm fails with target).
  • Transpose optimization to reduce time complexity when matrix is rectangular.
  • Time and space complexity analysis and comparison with brute force.

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