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Netflix·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026Remote

Summary

Netflix coding screen, one algorithmic question the whole time. The problem looked deceptively simple until you actually had to think about scaling it past brute force.

Questions Asked (1)

Q1

Given a list of lowercase show name strings, count the number of unique unordered pairs where the two strings share no common character.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Started with the obvious O(n^2) loop comparing every pair character by character, which they accepted but clearly wanted more.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints (e.g., input size, character set) and then propose an efficient algorithm. A common approach is to represent each string as a bitmask of its characters, then count pairs whose bitwise AND is zero. Discuss time and space complexity and possible optimizations.

Pro tip: Mention that since strings are lowercase, a 26-bit integer suffices, enabling fast bitwise operations. Also, consider precomputing masks and using a frequency map to handle duplicates efficiently.

1. Clarify requirements and constraints

Ask about input size, whether strings can be empty, and if duplicates are allowed. This determines the optimal approach.

2. Choose a representation

Represent each string as a bitmask where each bit indicates presence of a character. This reduces the problem to counting pairs with disjoint masks.

3. Design the counting algorithm

Use a frequency map of masks. For each mask, iterate over all possible masks (or use subset enumeration) to find disjoint ones, or use a hash map to count pairs efficiently.

4. Analyze complexity and trade-offs

Discuss time and space complexity. Compare brute-force O(n^2) with bitmask approach O(n * 2^26) or O(n * 2^26) is too large; instead, use a map and iterate over subsets of complement.

5. Handle edge cases and test

Consider empty strings, strings with all unique characters, and large inputs. Walk through a small example to verify correctness.

Key Points to Mention

  • Bitmask representation of character sets (26 bits for lowercase letters)
  • Using a frequency map to group identical masks
  • Counting disjoint pairs via bitwise AND == 0
  • Time complexity: O(n + m * 2^k) where m is number of unique masks and k is alphabet size, or O(n * 2^26) is too large; better to iterate over subsets of complement
  • Space complexity: O(m) for frequency map
  • Trade-offs: brute-force vs. bitmask optimization, and handling large inputs

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