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Meta·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Apr 2026Remote

Summary

Meta SWE phone screen that basically turned into a grid DP session with a sneaky robot-maze follow-up hiding in the back pocket. Two variants of the same core problem, and the gap between them is bigger than it looks on paper.

Questions Asked (3)

Q1

Given an m x n grid where each cell holds a reward value (which can be zero or negative), find the maximum total reward you can collect along a path from the top-left to the bottom-right corner, moving only right or down.

Algorithms & Data Structures
Author's notes

Pretty standard once you see it.

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AI HintsAI Generated

Suggested Approach

Clarify that this is a dynamic programming problem where you compute the maximum reward to reach each cell from the top-left, considering only moves from the top or left. Then present an O(m*n) time and O(n) space solution, explaining how to handle negative values and edge cases.

Pro tip: Emphasize that negative rewards are allowed, so you cannot greedily choose the locally larger neighbor; you must consider all paths. Also, mention that you can optimize space to O(n) by keeping only the previous row, which shows strong DP optimization skills.

1. Clarify the problem and constraints

Confirm that the grid can contain negative values, that movement is only right or down, and that you start at (0,0) and end at (m-1,n-1). Ask about grid size limits to determine if O(m*n) is acceptable.

2. Define the DP state and recurrence

Let dp[i][j] be the maximum reward to reach cell (i,j). Then dp[i][j] = grid[i][j] + max(dp[i-1][j], dp[i][j-1]), with base cases for the first row and first column.

3. Walk through a small example

Use a 2x2 or 3x3 grid with negative values to demonstrate how the DP table is filled and why greedy fails. Show the final answer at dp[m-1][n-1].

4. Analyze complexity and optimize space

State that time complexity is O(m*n) and space can be reduced from O(m*n) to O(n) by keeping only the previous row (or O(min(m,n)) by choosing the smaller dimension).

5. Discuss edge cases and potential follow-ups

Mention handling of 1x1 grid, grids with all negative values, and large grids. Be prepared to discuss variations like allowing all four directions or obstacles.

Key Points to Mention

  • Dynamic programming with optimal substructure and overlapping subproblems
  • Recurrence relation: dp[i][j] = grid[i][j] + max(dp[i-1][j], dp[i][j-1])
  • Base cases: first row and first column initialized by cumulative sums
  • Time complexity O(m*n) and space optimization to O(n)
  • Negative values require considering all paths, not just greedy choices
  • Edge cases: 1x1 grid, all negative values, large grids

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

Q2

Same maze/reward problem, but now you only have move() and canMove() API calls and a isCheese() check. No global grid view. How do you find the goal and maximize reward collected?

Algorithms & Data StructuresAPI & Integrations
Author's notes

This is where I stumbled.

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AI HintsAI Generated

Suggested Approach

Treat the maze as an unknown graph and use a systematic exploration algorithm like DFS or BFS with backtracking, leveraging only the provided APIs. To maximize reward, prioritize exploring branches that may contain cheese, and once the goal is found, compute an optimal path that collects as much reward as possible.

Pro tip: Explicitly discuss the trade-off between exploration and exploitation: sometimes it's worth taking a longer path to collect more cheese, but you must ensure you can still reach the goal. Also, mention that you'd cache the maze structure as you explore to avoid redundant moves.

1. Clarify API semantics and constraints

Ask about move() behavior (does it move in a direction and return success?), canMove() (checks if a move is possible without moving?), and isCheese() (checks current cell?). Confirm if moves are reversible and if there's a limit on moves.

2. Choose an exploration strategy

Use DFS with backtracking to explore the maze systematically, or BFS if you want to find the shortest path to the goal first. Since there's no global view, you'll need to remember visited cells and paths.

3. Handle reward collection

When you detect cheese (via isCheese()), decide whether to collect it immediately or mark it for later. If collecting, ensure you can return to the main path. Consider using a priority system to visit cheese-rich areas first.

4. Optimize for reward and goal

Once the goal is found, you may need to backtrack to collect missed cheese. Alternatively, during exploration, keep track of all cheese locations and plan a route that maximizes reward while still reaching the goal.

5. Analyze complexity and edge cases

Discuss time and space complexity in terms of maze size and number of cheese. Address edge cases like unreachable cheese, cycles, and the possibility of infinite loops if not careful.

Key Points to Mention

  • Use of DFS/BFS with backtracking to explore unknown environment
  • Maintaining a visited set or map to avoid cycles and redundant moves
  • Strategy for collecting cheese: greedy vs. optimal path planning
  • Trade-off between exploration and exploitation (reward vs. goal)
  • Complexity analysis: O(V+E) for graph traversal, where V is cells and E is connections
  • Handling of edge cases: no cheese, cheese behind dead ends, goal unreachable

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

Q3

Follow-up: what if the robot can now move in all four directions instead of just right and down? Does your DP approach still work?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

DP breaks because you can revisit cells and the subproblem ordering falls apart, especially with negative rewards.

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AI HintsAI Generated

Suggested Approach

Acknowledge that the DP approach for counting paths with only right/down moves does not directly extend to four-direction movement because cycles are introduced, making the problem about shortest paths or reachability rather than simple path counting. Then pivot to discussing appropriate algorithms like BFS for unweighted grids or Dijkstra for weighted grids, and mention how obstacles or constraints affect the choice.

Pro tip: Show awareness that the original DP relied on a DAG (no cycles), and that adding up/left moves breaks that assumption—demonstrating you understand the underlying reason, not just the algorithm. Also, briefly mention that if the goal is counting simple paths (no revisits), the problem becomes #P-complete, which is a great way to show depth.

1. Clarify the problem

Ask whether the goal is still to count all paths, find the shortest path, or just determine reachability. Also confirm if revisiting cells is allowed and if there are obstacles or weights.

2. Explain why DP fails

State that the original DP works because moves only go right/down, forming a DAG with a topological order. With four directions, cycles exist, so DP over cells without additional state (like visited set) is invalid.

3. Propose alternative algorithms

For shortest path in an unweighted grid, use BFS. For weighted grids, use Dijkstra. For counting simple paths, note it's #P-complete and likely infeasible for large grids.

4. Discuss trade-offs and constraints

Compare BFS vs. Dijkstra vs. A* based on grid size, weights, and whether we need all shortest paths or just one. Mention that if the grid is small, DFS with backtracking can count simple paths.

5. Summarize and connect to original

Conclude that the DP approach no longer applies directly, but the problem transforms into a classic graph search, and the choice depends on the exact requirement.

Key Points to Mention

  • DP relies on acyclic dependencies (DAG) from right/down moves; four directions introduce cycles.
  • BFS is optimal for unweighted shortest path; Dijkstra for weighted.
  • Counting simple paths in a grid with cycles is #P-complete (or NP-hard for decision version).
  • If revisiting is allowed, infinite paths exist unless we restrict to simple paths.
  • Obstacles or weights change the algorithm choice (e.g., A* with heuristics).
  • State-space search with memoization on (cell, visited set) is exponential and impractical for large grids.

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