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Meta·Software Engineer·Onsite - Coding / Algorithms·Intermediate

IntermediatePrefer not to say
May 2026

Summary

Meta SWE coding round, one problem the whole time. Pretty standard anagram grouping question but I still managed to second-guess myself halfway through.

Questions Asked (1)

Q1

Given an array of strings, group all anagrams together and return the groups. Two strings are anagrams if they have the same characters at the same frequencies.

Algorithms & Data Structures
Author's notes

Knew the sorted-key trick going in but still spent like two minutes overthinking whether a hash of character counts would be faster.

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AI HintsAI Generated

Suggested Approach

Use a hash map to group strings by a canonical key that represents their character frequencies. For each string, compute the key (e.g., sorted string or frequency tuple) and append the string to the corresponding group. Finally, return all groups as a list of lists.

Pro tip: Discuss trade-offs between sorting each string (O(n * k log k)) and using a frequency count key (O(n * k)), and mention that the frequency approach is more efficient for large alphabets or long strings. Also, clarify that the order of groups and strings within groups doesn't matter unless specified.

1. Clarify requirements and constraints

Ask about input size, character set (e.g., lowercase letters only?), and whether the output order matters. This shows attention to detail and helps choose the optimal approach.

2. Choose a canonical key

Decide on a key that uniquely identifies anagrams: either the sorted string or a frequency count (e.g., a tuple of 26 counts for lowercase letters). Explain why the key works.

3. Iterate and group

Initialize a hash map. For each string, compute its key and append the string to the list associated with that key in the map.

4. Return the groups

Extract all values from the hash map and return them as a list of lists. Mention that the order of groups is arbitrary.

5. Analyze complexity

State the time and space complexity: O(n * k) for frequency key (where n is number of strings, k is max length) or O(n * k log k) for sorting, and O(n * k) space for storing the groups.

Key Points to Mention

  • Hash map (dictionary) for grouping by key
  • Canonical key: sorted string vs. frequency count
  • Time complexity: O(n * k) with frequency count, O(n * k log k) with sorting
  • Space complexity: O(n * k) for storing all strings
  • Handling edge cases: empty strings, single string, all anagrams
  • Trade-offs between approaches and when to use each

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