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Airbnb·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Airbnb SWE interview that hit me with two pretty tough LeetCode problems back to back. Not much context given about the round itself but the problems speak for themselves.

Questions Asked (2)

Q1

Given a list of words, find all pairs of indices where concatenating the two words forms a palindrome.

Algorithms & Data Structures
Author's notes

This one is rough.

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AI HintsAI Generated

Suggested Approach

Use a hash map to store each word along with its index, then for each word, split it at every possible position and check if the prefix/suffix can form a palindrome with another word. Handle edge cases like empty strings and duplicate words carefully.

Pro tip: Clarify whether the two words can be the same index (i.e., i != j) and whether the order matters (i.e., (i, j) and (j, i) are distinct). Also, discuss time complexity trade-offs between brute force and optimized approach.

1. Understand the problem and edge cases

Confirm that pairs must have distinct indices (i != j) and that order matters (i, j) and (j, i) are different. Consider empty strings, duplicate words, and words that are themselves palindromes.

2. Brute force approach

Iterate over all pairs (i, j) with i != j, concatenate words[i] + words[j], and check if the result is a palindrome. This is O(n^2 * L) where L is the average word length.

3. Optimized approach using hash map

Store each word and its index in a hash map. For each word, split it into two parts at every position: prefix and suffix. Check if the prefix is a palindrome and if the reverse of the suffix exists in the map (and is not the same index). Similarly, check if the suffix is a palindrome and if the reverse of the prefix exists.

4. Handle duplicates and empty strings

When checking for the reverse of a part, ensure the index is different. For empty strings, they can pair with any palindrome word, but be careful not to double-count.

5. Analyze complexity and test

The optimized approach runs in O(n * L^2) time and O(n * L) space. Walk through examples like ['bat', 'tab', 'cat'] and edge cases like ['', 'a'] to verify correctness.

Key Points to Mention

  • Use a hash map to store words and their indices for O(1) lookups.
  • Split each word at every possible position to check prefix/suffix palindrome conditions.
  • Ensure i != j and avoid duplicate pairs by checking indices.
  • Handle empty strings and words that are palindromes themselves.
  • Time complexity: O(n * L^2) where n is number of words and L is max word length.
  • Space complexity: O(n * L) for the hash map.

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

Q2

Given a grid with walls, open cells, and keys/locks, find the shortest path that collects all keys.

Algorithms & Data Structures
Author's notes

BFS with bitmask state.

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AI HintsAI Generated

Suggested Approach

Model the problem as a state-space search where each state includes the current position and the set of keys collected. Use BFS to find the shortest path, since all moves have equal cost. Represent the key set as a bitmask to efficiently track progress.

Pro tip: Before coding, clarify constraints like grid size and number of keys; if keys are few (≤10), bitmask BFS is optimal, but if many, consider A* with a heuristic like minimum spanning tree over remaining keys.

1. Clarify problem details

Ask about grid dimensions, number of keys, whether locks require specific keys, and if multiple keys can be collected. Confirm that you need the shortest path length, not the actual path.

2. Define state representation

Use a tuple (row, col, keys_bitmask) to represent the state. The bitmask tracks which keys have been collected, with each bit corresponding to a key.

3. Perform BFS with state tracking

Start BFS from the initial position with an empty key set. For each move, update the key set if a key is picked up, and only pass through a lock if the corresponding key is held. Use a visited set to avoid revisiting states.

4. Check termination and return result

When the key set contains all keys, return the current distance. If BFS exhausts all states without collecting all keys, return -1 or indicate no solution.

5. Analyze complexity and optimize

Discuss time complexity O(R*C*2^K) and space complexity O(R*C*2^K). Mention possible optimizations like bidirectional BFS or A* if needed.

Key Points to Mention

  • State-space search with BFS for shortest path in unweighted grid.
  • Bitmask representation for key collection to efficiently track subsets.
  • Handling locks: only traverse if the corresponding key is collected.
  • Visited set must include the key bitmask to avoid redundant states.
  • Time and space complexity analysis in terms of grid size and number of keys.
  • Edge cases: no keys, unreachable keys, multiple keys of same type, start on a key.

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