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Meta·Software Engineer·Technical Phone Screen·Senior

SeniorPrefer not to say
Apr 2026Remote

Summary

Meta SWE interview focused on a grid traversal problem that looked straightforward until the follow-ups started piling up. The whole session felt like one question that kept branching into five more.

Questions Asked (4)

Q1

Given an m by n grid of integers, find the length of the longest path where each consecutive step moves to an adjacent cell (up, down, left, or right) with a strictly larger value. Return the maximum over all possible starting cells.

Algorithms & Data Structures
Author's notes

I started with brute-force DFS and they let me talk through it before asking about complexity.

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AI HintsAI Generated

Suggested Approach

Model the grid as a directed acyclic graph where edges go from a cell to adjacent cells with strictly larger values. Use DFS with memoization to compute the longest increasing path starting from each cell, caching results to avoid redundant work. Return the maximum over all starting cells.

Pro tip: Emphasize that memoization is valid because the strictly increasing condition ensures no cycles, so each cell's longest path is independent of how you reached it. Also mention that you can avoid explicit graph construction by computing neighbors on the fly.

1. Clarify and Define

Confirm that the path can start at any cell and moves only to adjacent cells with strictly larger values. Clarify that diagonal moves are not allowed and that the path length is the number of cells visited.

2. Identify Optimal Substructure

Recognize that the longest path from a cell is 1 plus the maximum longest path from its valid neighbors. This recursive relationship allows dynamic programming.

3. Choose Algorithm

Use DFS with memoization (top-down DP) to compute the longest path from each cell. Alternatively, use topological sort on the DAG of cells ordered by value, but DFS+memo is simpler.

4. Implement and Optimize

Initialize a memo table with zeros. For each cell, if not computed, recursively compute the longest path by exploring up to four neighbors with larger values. Track the global maximum.

5. Analyze Complexity

Each cell is visited once, and each edge (up to 4 per cell) is considered once, giving O(m*n) time and O(m*n) space for memoization and recursion stack.

Key Points to Mention

  • The problem is equivalent to finding the longest path in a directed acyclic graph (DAG) where edges go from smaller to larger values.
  • Memoization ensures each cell's longest path is computed only once, reducing time complexity from exponential to linear.
  • The strictly increasing condition guarantees no cycles, so the graph is a DAG and DP is valid.
  • Space complexity includes the memoization table and the recursion stack, both O(m*n) in the worst case.
  • You can avoid building an explicit graph by checking neighbors on the fly, saving memory.
  • Edge cases: empty grid, single cell, all equal values (answer 1), and large grids where recursion depth might be an issue (consider iterative DP or increasing recursion limit).

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

Q2

Walk through the time and space complexity of your memoized DFS solution for the longest increasing path problem.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Said O(mn) time and O(mn) space and that was basically it.

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AI HintsAI Generated

Suggested Approach

Start by clearly defining the problem and the memoized DFS approach, then systematically derive time and space complexity by analyzing the number of states and transitions. Conclude by discussing trade-offs and potential optimizations.

Pro tip: Explicitly state that each cell is computed once and each edge is traversed once, leading to O(mn) time, and that the memoization table and recursion stack contribute to O(mn) space. This shows you understand the amortized analysis and can communicate it concisely.

1. Define the problem and approach

Briefly restate the longest increasing path problem and explain that memoized DFS computes the longest path starting from each cell, caching results to avoid redundant work.

2. Analyze time complexity

Argue that each cell is visited once as a starting point, and for each cell, we explore up to four neighbors. Since memoization ensures each cell's result is computed only once, the total work is proportional to the number of cells plus the number of edges, yielding O(mn) time.

3. Analyze space complexity

The memoization table stores one value per cell, taking O(mn) space. Additionally, the recursion stack can go up to O(mn) in the worst case (e.g., a strictly increasing path), so total space is O(mn).

4. Discuss trade-offs and optimizations

Mention that while O(mn) is optimal for this problem, iterative DP with topological sort can avoid recursion overhead, and that space can be reduced if we only need the length (but not the path) by using a 1D array if processing in a specific order.

Key Points to Mention

  • Each cell is computed once due to memoization, leading to O(mn) time.
  • Each cell has at most 4 neighbors, so total edge traversals are O(mn).
  • Memoization table uses O(mn) space.
  • Recursion stack depth can be O(mn) in the worst case (e.g., a snake-like increasing path).
  • Time and space complexity are both O(mn), which is optimal for this problem.
  • Alternative iterative DP with topological sort can achieve the same complexity but with different constant factors.

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

Q3

How would you handle this problem on very large grids where recursion depth could become a bottleneck or hit stack limits?

Algorithms & Data StructuresTechnical Trade-offsSystem Design
Author's notes

This is where I got a bit shaky.

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AI HintsAI Generated

Suggested Approach

Acknowledge the recursion depth issue and propose converting the recursive solution to an iterative one using an explicit stack or queue. Discuss trade-offs between BFS and DFS, and mention memory considerations for very large grids. Emphasize that the iterative approach avoids stack overflow and can be more memory-efficient if implemented carefully.

Pro tip: Mention that you can also use a hybrid approach: recursion with increased stack size or tail-call optimization where supported, but iterative is generally safer and more portable. Also, highlight the importance of early termination and pruning to reduce the search space.

1. Identify the problem

Recognize that recursion depth on large grids can lead to stack overflow due to limited call stack size.

2. Choose an iterative approach

Convert the recursive algorithm to an iterative one using an explicit data structure like a stack (for DFS) or queue (for BFS).

3. Analyze trade-offs

Compare BFS vs DFS in terms of memory usage, time complexity, and suitability for the problem (e.g., shortest path vs exhaustive search).

4. Optimize memory

Consider using a compact representation for visited cells (e.g., bitset) and avoid storing unnecessary data in the stack/queue.

5. Handle edge cases

Ensure the iterative solution handles large grids without excessive memory usage and includes early termination conditions.

Key Points to Mention

  • Stack overflow risk with deep recursion
  • Iterative DFS using explicit stack
  • BFS with queue for shortest path
  • Memory trade-offs: stack vs queue size
  • Visited set optimization (e.g., in-place marking or bitset)
  • Tail recursion and compiler optimizations (if applicable)

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

Q4

What changes if the path condition is non-decreasing instead of strictly increasing? What about allowing diagonal moves?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The non-decreasing variant breaks the DAG property since you can have cycles between equal values, which means memoization alone doesn't save you.

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AI HintsAI Generated

Suggested Approach

First, clarify the problem context—likely a grid DP problem like counting paths with moves right/down. Then, analyze how each change affects the state transitions and base cases: non-decreasing path condition allows equal values, introducing dependencies on equal-valued cells; diagonal moves add a third direction, increasing branching. Finally, discuss the impact on time/space complexity and potential optimizations.

Pro tip: Always relate the changes to the underlying DP recurrence and consider edge cases like all cells equal or obstacles. Mentioning how to handle cycles or dependencies shows depth.

1. Clarify the problem and assumptions

Confirm the original problem: likely counting paths in a grid with strictly increasing values and moves right/down. State assumptions about grid size, value range, and obstacles.

2. Analyze non-decreasing condition

Explain that non-decreasing allows equal values, so paths can include cells with the same value. This may introduce dependencies among equal-valued cells, requiring careful ordering (e.g., process by value groups) to avoid cycles.

3. Analyze diagonal moves

Adding diagonal moves increases the number of transitions per cell from 2 to 3 (or more if all diagonals allowed). This changes the DP recurrence and may increase time complexity by a constant factor, but could also enable new paths.

4. Combine both changes

Consider the combined effect: non-decreasing with diagonal moves. Discuss how to handle equal values with additional move directions, and whether the problem becomes more complex (e.g., need for topological sort or union-find).

5. Discuss complexity and optimizations

Compare time/space complexity of the variants. Mention potential optimizations like sorting cells by value, using BFS/DFS with memoization, or leveraging union-find for equal-value groups.

Key Points to Mention

  • DP state definition: dp[i][j] = number of valid paths to (i,j).
  • Transition: from top and left (and diagonal if allowed).
  • Non-decreasing allows equal values, so dp[i][j] may depend on dp[i-1][j] and dp[i][j-1] even if values equal, but careful with cycles if moves can go to equal values in multiple directions.
  • Diagonal moves add dp[i-1][j-1] to the transition.
  • Complexity: O(m*n) for grid DP, but with non-decreasing and equal values, may need to process cells in order of value, potentially O(m*n log(m*n)) if sorting.
  • Edge cases: all cells equal, obstacles, large grids, and value ranges.

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