Meta·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jul 2026

Summary

Meta SWE coding round, one problem the whole session. Clean premise but the edge cases sneak up on you if you're not careful.

Questions Asked (1)

Given a list of lowercase English words, find the largest subset where every character across all chosen words appears exactly once. Words with internal duplicate letters are disqualified. Return the total count of unique characters covered.

Algorithms & Data Structures

Author's notes

My first instinct was greedy but that falls apart fast.

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AI HintsAI Generated

Suggested Approach

First, filter out words with duplicate letters and represent each valid word as a bitmask of its characters. Then, use backtracking or dynamic programming to find the maximum number of unique characters covered by a subset of these masks with no overlapping bits.

Pro tip: Clarify that the goal is to maximize the total count of unique characters, not the number of words. Also, mention that bitmask DP can be optimized by grouping words with the same mask and only keeping the best (though all have the same character count).

1. Validate and Encode Words

Iterate through each word, check for duplicate characters, and if valid, compute a bitmask representing the set of characters in the word.

2. Choose an Algorithm

Decide between backtracking (exploring all subsets) and dynamic programming (using a map from mask to max characters) based on constraints and desired efficiency.

3. Implement the Solution

If backtracking, recursively try including or excluding each word, ensuring no character overlap. If DP, iterate through words and update a dictionary of achievable masks.

4. Track and Return the Maximum

Maintain the maximum number of unique characters seen so far and return it after processing all words.

Key Points to Mention

Bitmask representation of character sets for efficient overlap checking.
Filtering out words with duplicate letters as they can never be part of a valid subset.
The problem is equivalent to finding a maximum weight independent set in a conflict graph, but bitmask DP exploits the small alphabet size.
Time complexity: O(N * 2^26) in worst case for DP, but practically much less due to pruning; backtracking can be exponential but often fast with pruning.
Space complexity: O(2^26) for DP if using an array, but a hash map reduces it to the number of reachable masks.
Edge cases: empty list, all words invalid, words with no overlapping characters.

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