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Airbnb·Software Engineer·Technical Phone Screen·Senior

Senior
May 2026

Summary

Airbnb software engineering interview with a pretty gnarly algorithmic problem centered on palindrome pair detection at scale. The question had a lot of moving parts and felt like it was testing whether you'd actually thought about trie-based string processing before.

Questions Asked (1)

Q1

Given an array of distinct lowercase strings, find all index pairs (i, j) where i != j and the concatenation of words[i] + words[j] forms a palindrome. The array can have up to 100,000 words with total character count up to 200,000. You need to beat the naive O(n^2 * L) approach. Walk through your data structure choices, how you handle edge cases like empty strings and single characters, how you prevent duplicate pairs, and give a complexity analysis with test cases.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This wrecked me a little.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and constraints, then propose a hash map-based solution that stores reversed words and checks for palindromic splits, handling edge cases like empty strings and duplicates. Walk through the algorithm step-by-step, analyze time and space complexity, and discuss test cases including large inputs.

Pro tip: Emphasize that the total character count is 200,000, so an O(total characters) solution is feasible; mention that using a trie can optimize further but a hash map is simpler and sufficient. Also, proactively discuss how to avoid duplicate pairs by enforcing i < j or using a set.

1. Clarify and Restate

Confirm understanding of the problem: distinct lowercase strings, find all index pairs (i, j) with i != j such that words[i] + words[j] is a palindrome. Note constraints: up to 100,000 words, total characters 200,000, need better than O(n^2 * L).

2. Design Data Structures

Use a hash map to store each word's index for O(1) lookups. For each word, consider all possible splits into prefix and suffix, check if one part is a palindrome and the other's reverse exists in the map.

3. Handle Edge Cases

Address empty strings: if a word is empty, it can pair with any palindrome word. Single characters are palindromes. Ensure no duplicate pairs by only adding when i != j and using a set or checking indices.

4. Complexity Analysis

Time: O(N * L^2) worst-case if checking each split naively, but with precomputed palindrome checks or efficient methods, it can be O(total characters * average word length) or O(N * L) with optimizations. Space: O(N * L) for the hash map.

5. Test Cases

Include: empty array, single word, words with empty string, words like 'a', 'ab', 'ba', 'abc', 'cba', and large random inputs to verify performance. Check for duplicate pairs and correct palindrome formation.

Key Points to Mention

  • Hash map for O(1) lookups of reversed words.
  • Splitting each word into all possible prefix-suffix pairs and checking palindrome conditions.
  • Handling empty strings and single-character words as special cases.
  • Avoiding duplicate pairs by enforcing i < j or using a set.
  • Time complexity: O(N * L^2) worst-case, but can be optimized to O(N * L) with precomputed palindrome checks.
  • Space complexity: O(N * L) for storing words and indices.

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