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Pilot·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Did a technical screen for a Software Engineer role at Pilot. One coding question, pretty focused, no fluff. The constraint about not sorting threw me for a loop at first.

Questions Asked (1)

Q1

Given two strings, determine whether one is an anagram of the other. Sorting is not allowed. Aim for a linear time solution.

Algorithms & Data Structures
Author's notes

The no-sorting constraint is what makes this interesting.

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AI HintsAI Generated

Suggested Approach

Clarify assumptions (case sensitivity, character set, whitespace) and edge cases (length mismatch, empty strings). Then propose a frequency count using a hash map or fixed-size array, achieving O(n) time and O(1) or O(k) space. Walk through the algorithm, analyze complexity, and discuss trade-offs versus sorting.

Pro tip: Mention that if the character set is fixed (e.g., ASCII), a 256-element array gives O(1) space and faster constant factors than a hash map. Also, note that early length check avoids unnecessary work.

1. Clarify requirements and edge cases

Ask about case sensitivity, allowed characters, whitespace, and Unicode. Confirm that sorting is disallowed and linear time is required. Check edge cases like empty strings or different lengths.

2. Propose frequency counting approach

Explain that you will count character frequencies in the first string and decrement for the second. Use a hash map for general characters or a fixed-size array for a known character set.

3. Walk through the algorithm

Describe initializing the count structure, iterating through the first string to increment counts, then iterating through the second to decrement. If any count goes negative or final counts are non-zero, return false.

4. Analyze time and space complexity

State that time is O(n) where n is the length of the strings, and space is O(k) where k is the number of distinct characters (or O(1) for fixed alphabet). Compare with sorting's O(n log n).

5. Discuss trade-offs and optimizations

Mention that for small strings, sorting might be simpler but not linear. For fixed alphabets, an array is more efficient. Also, note that early length check can quickly reject non-anagrams.

Key Points to Mention

  • Length check as a quick rejection
  • Hash map vs. fixed-size array for frequency counting
  • Time complexity O(n) and space complexity O(k) or O(1)
  • Handling Unicode or extended character sets
  • Early termination if a count becomes negative
  • Comparison with sorting approach (O(n log n))

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