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Airbnb·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Airbnb SWE interview with a grid-based algorithm problem that started straightforward and then got a lot harder with the follow-up. The single-skier version felt manageable but the multi-skier extension is where things got real.

Questions Asked (2)

Q1

You're given a 2D grid of integers representing elevations. A skier starts at a given cell and can move up, down, left, or right, but only to a cell with strictly lower elevation. What's the maximum number of cells the skier can visit starting from a given position, including the start?

Algorithms & Data Structures
Author's notes

My first instinct was DFS with memoization and that ended up being the right call.

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AI HintsAI Generated

Suggested Approach

Model the grid as a directed acyclic graph where edges go from higher to lower elevation, then use DFS with memoization to compute the longest path from each cell. Start from the given position and return the maximum length found.

Pro tip: Clarify whether the skier can revisit cells or if the path must be simple; the strictly lower elevation condition guarantees no cycles, so memoization is safe. Also, mention that you can optimize space by using a 2D array for memoization and early termination if the current path length plus remaining possible cells cannot exceed the best found.

1. Clarify the problem

Confirm that the skier can only move to strictly lower elevations, that diagonal moves are not allowed, and that the path length includes the starting cell. Ask if the grid can be modified or if extra space is allowed.

2. Define the recursive relation

For a cell (i, j), the maximum path length is 1 + max of the lengths from all valid neighbors with lower elevation. If no such neighbor exists, the length is 1.

3. Choose memoization or DP

Use a memoization table (e.g., 2D array) to store computed lengths for each cell to avoid redundant work. Alternatively, sort cells by elevation and compute lengths in increasing order (dynamic programming).

4. Implement DFS with memoization

Write a recursive function that checks the memo table first, then explores all four directions, updating the maximum length. Ensure to mark cells as visited or use the memo table to avoid recomputation.

5. Analyze complexity and edge cases

Time complexity is O(m*n) because each cell is computed once. Space complexity is O(m*n) for the memo table. Discuss edge cases like a single cell, all equal elevations, or a strictly decreasing grid.

Key Points to Mention

  • Directed acyclic graph (DAG) representation: edges from higher to lower elevation.
  • Memoization to avoid recomputing overlapping subproblems.
  • Time and space complexity: O(m*n) time, O(m*n) space.
  • Handling of boundary conditions and invalid moves.
  • Comparison with alternative approaches like topological sort or sorting cells by elevation.
  • Potential optimization: iterative DP with sorting to avoid recursion depth issues.

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

Q2

Follow-up: if you have many skiers each with their own starting cell, how do you return the score for each one efficiently without recomputing from scratch every time?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where I started sweating.

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AI HintsAI Generated

Suggested Approach

Recognize this as a multi-source dynamic programming problem where each starting cell is a source. Precompute the score for all cells in a single pass using a multi-source BFS/DFS (or topological order if DAG), then answer each query in O(1) by lookup. Emphasize that this avoids redundant computation and scales well with many skiers.

Pro tip: Mention that you can further optimize by grouping skiers with the same starting cell or using memoization with a shared cache, and discuss trade-offs like memory vs. time. This shows you think about practical constraints and not just the algorithm.

1. Clarify the problem and constraints

Confirm the definition of 'score' (e.g., longest increasing path, max sum, etc.) and whether the grid is static. Ask about the number of skiers, grid size, and if multiple skiers can start at the same cell.

2. Identify the core algorithmic pattern

Recognize that computing scores from many starting points is equivalent to a multi-source traversal. If the score is based on longest path in a DAG (e.g., strictly decreasing heights), use topological sort or DFS with memoization to compute all scores in one pass.

3. Design the precomputation

Initialize a score array. For multi-source BFS/DFS, enqueue all starting cells with their initial scores, then propagate. For DAG, compute in topological order from sinks to sources. Ensure each cell's score is computed once.

4. Answer queries efficiently

After precomputation, each skier's score is simply the precomputed value at their starting cell. This gives O(1) per query after O(V+E) preprocessing.

5. Discuss trade-offs and optimizations

Compare with recomputing per skier (O(K*(V+E)) vs O(V+E + K)). Mention memory usage, potential for parallelization, and handling dynamic updates if needed.

Key Points to Mention

  • Multi-source BFS/DFS or topological sort for DAGs
  • Memoization/dynamic programming to avoid recomputation
  • Time complexity: O(V+E) preprocessing, O(1) per query
  • Space complexity: O(V) for score array
  • Handling multiple skiers at same starting cell (cache reuse)
  • Trade-offs: preprocessing cost vs. query cost, memory vs. time

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