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Meta·Machine Learning Engineer·Technical Phone Screen·Senior

Senior
Apr 2026

Summary

Meta ML engineer interview with a coding question that looked like a string problem but was really about understanding rotational equivalence. Pretty lean on details but the core puzzle was interesting enough.

Questions Asked (1)

Q1

Given a list of strings, group together all strings that are equivalent under some Caesar Cipher rotation (where every character is shifted by the same fixed amount, wrapping around the alphabet).

Algorithms & Data Structures
Author's notes

My first instinct was to just try all 26 rotations for each string and use that as a key, which works but felt clunky.

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AI HintsAI Generated

Suggested Approach

Normalize each string to a canonical form by shifting it so that its first character becomes 'a', then group strings by this canonical key using a hash map. This reduces the problem to finding a consistent representative for each equivalence class under Caesar shifts.

Pro tip: Clarify whether the shift is applied uniformly to all characters and whether the alphabet is circular; also discuss handling empty strings and strings with non-alphabetic characters, as these edge cases often trip candidates.

1. Understand the problem and constraints

Confirm that two strings are equivalent if one can be obtained from the other by shifting every character by the same amount (mod 26). Ask about alphabet size, character set, and empty strings.

2. Design a canonical representation

For each string, compute the shift needed to make its first character 'a' (or any fixed character), then apply that shift to all characters to get a normalized string. This normalized string uniquely identifies the equivalence class.

3. Group using a hash map

Use a dictionary mapping the canonical string to a list of original strings. Iterate through the input list, compute the canonical form, and append the original string to the corresponding list.

4. Analyze complexity and edge cases

Time complexity is O(N * L) where N is number of strings and L is average length. Space is O(N * L). Discuss handling of empty strings, strings of different lengths, and non-alphabetic characters.

5. Test and validate

Walk through examples like ['abc', 'bcd', 'xyz'] to ensure they group correctly. Consider edge cases such as empty strings, single-character strings, and strings with uppercase letters.

Key Points to Mention

  • Canonical form: shift each string so its first character becomes 'a' (or any fixed reference).
  • Hash map for grouping: key is canonical string, value is list of original strings.
  • Time and space complexity: O(N * L) time, O(N * L) space.
  • Edge cases: empty strings, strings of different lengths, non-alphabetic characters, uppercase vs lowercase.
  • Alternative: use a tuple of differences between consecutive characters as the key.
  • Modular arithmetic for wrapping around the alphabet (mod 26).

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