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Bloomberg·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026

Summary

Bloomberg SWE interview that leaned heavily on graph traversal. The whole session was basically one problem stretched across five angles, which I didn't expect. Left feeling okay about the core solution but shaky on some of the follow-ups.

Questions Asked (5)

Q1

Given an m x n grid of '1's (land) and '0's (water), count the number of distinct islands where connectivity is 4-directional only. Implement a solution running in O(m*n) time and explain why it's correct.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I went with DFS and flood-fill, which felt natural.

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AI HintsAI Generated

Suggested Approach

Use a graph traversal algorithm like BFS or DFS to explore each unvisited land cell, marking all connected land cells as visited to count one island. Ensure each cell is visited at most once to achieve O(m*n) time, and explain that the traversal covers all cells and correctly identifies connected components.

Pro tip: Mention that you can optimize space by modifying the grid in-place (e.g., changing '1' to '0') to mark visited cells, avoiding a separate visited matrix. Also, clarify that 4-directional connectivity means only up, down, left, right neighbors are considered, which is crucial for correctness.

1. Clarify the problem and constraints

Restate the problem: count distinct islands in a binary grid with 4-directional connectivity. Confirm that islands are separate if not connected orthogonally, and that the grid can be modified or we can use extra space.

2. Choose traversal algorithm

Select BFS or DFS for exploring connected components. Explain that either works, but BFS avoids recursion depth issues for large grids, while DFS is simpler to code.

3. Implement traversal with visited marking

Iterate through each cell; when encountering an unvisited '1', increment island count and launch traversal to mark all connected '1's as visited (e.g., by setting to '0' or using a visited set).

4. Analyze time and space complexity

Argue that each cell is visited at most once, so time is O(m*n). Space is O(min(m,n)) for BFS queue or O(m*n) worst-case for DFS recursion, but can be O(1) if modifying grid in-place.

5. Prove correctness

Explain that the algorithm counts exactly one island per connected component because traversal from an unvisited land cell marks all reachable land cells, and no cell is counted twice.

Key Points to Mention

  • 4-directional connectivity (up, down, left, right) and why diagonal connections don't count.
  • Using BFS with a queue or DFS with recursion/stack for traversal.
  • Marking visited cells to avoid infinite loops and double-counting.
  • Time complexity O(m*n) because each cell is processed once.
  • Space complexity considerations and possible in-place modification to achieve O(1) extra space.
  • Edge cases: empty grid, all water, all land, single row/column.

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

Q2

Re-implement the island count using an explicit BFS with an iterative queue instead of recursion. How does the complexity compare to your DFS solution?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Honestly the easier part for me.

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AI HintsAI Generated

Suggested Approach

Walk through the BFS implementation by replacing the recursive DFS call with an explicit queue, processing each cell's neighbors iteratively. Then compare the time and space complexity of both approaches, noting that while both are O(M*N) time, BFS uses O(min(M,N)) space in the worst case due to the queue, whereas DFS uses O(M*N) space from the call stack in the worst case. Emphasize that the choice depends on constraints and that BFS avoids recursion depth limits.

Pro tip: Mention that BFS can be more memory-efficient for large grids because the queue size is bounded by the perimeter of the island, but in the worst case (e.g., a spiral-shaped island) it can still grow to O(M*N). Also, note that BFS explores level by level, which might be useful if you need to find the shortest path or distance from the starting point.

1. Outline BFS approach

Explain that you'll iterate over each cell; when you find a '1', increment the island count and start a BFS using a queue to mark all connected land cells as visited.

2. Detail queue operations

Describe initializing the queue with the starting cell, then while the queue is not empty, dequeue a cell and enqueue all valid unvisited neighboring land cells, marking them visited immediately to avoid duplicates.

3. Analyze time complexity

State that both BFS and DFS visit each cell at most once, so time complexity is O(M*N) where M and N are grid dimensions.

4. Compare space complexity

Explain that DFS uses O(M*N) space in the worst case due to recursion stack, while BFS uses O(min(M,N)) space in the average case because the queue holds at most the perimeter of the island, but worst-case can still be O(M*N).

5. Discuss trade-offs

Mention that BFS avoids stack overflow for large grids, but may use more memory in some cases; DFS is simpler to code recursively but risks recursion depth limits.

Key Points to Mention

  • Time complexity remains O(M*N) for both BFS and DFS.
  • Space complexity: DFS O(M*N) worst-case due to call stack; BFS O(min(M,N)) average-case due to queue size bounded by island perimeter.
  • BFS uses an explicit queue, eliminating recursion depth concerns.
  • Mark cells as visited when enqueuing to prevent duplicate processing.
  • BFS explores level by level, which can be advantageous for shortest path problems.
  • In the worst case (e.g., spiral island), BFS queue can also grow to O(M*N).

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

Q3

How does your solution handle edge cases like a null grid, all-water grid, single-cell grid, fully-land grid, very large grids, and diagonal-only touching cells?

Algorithms & Data Structures
Author's notes

Ran through these pretty quickly.

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AI HintsAI Generated

Suggested Approach

Walk through each edge case systematically, explaining how your solution handles it without crashing or producing incorrect results. Emphasize the importance of early returns for trivial cases and the use of iterative traversal to avoid stack overflow on large grids. Clarify that diagonal touching is not considered connected, so your algorithm only checks 4-directional adjacency.

Pro tip: Mention that you would write unit tests for each edge case to ensure robustness, and discuss the time and space complexity implications for large grids, showing you think about scalability.

1. Acknowledge and enumerate edge cases

List all mentioned edge cases and briefly explain why each is important to handle. This shows thoroughness and awareness of potential pitfalls.

2. Explain handling of trivial cases

Describe how your solution checks for null or empty grid, single-cell grid, all-water, and fully-land grids, often with early returns or simple checks.

3. Address large grids

Discuss using iterative BFS/DFS or union-find to avoid recursion depth issues, and mention memory considerations and optimizations like in-place modification.

4. Clarify connectivity definition

State that diagonal touching is not considered connected, so only 4-directional neighbors are explored. This prevents misinterpretation of the problem.

5. Summarize complexity and testing

Conclude with time/space complexity for each case and mention writing unit tests to validate edge case handling.

Key Points to Mention

  • Null or empty grid: return 0 immediately.
  • Single-cell grid: check if land, return 1 if so, else 0.
  • All-water grid: return 0 without traversal.
  • Fully-land grid: return 1 (one island) if connectivity is 4-directional.
  • Large grids: use iterative BFS/DFS or union-find to avoid stack overflow; consider memory usage.
  • Diagonal-only touching cells: treat as separate islands; only check up, down, left, right.

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

Q4

Describe or write tests that validate your solution on several nontrivial inputs, including the 3-island example from the problem.

Algorithms & Data Structures
Author's notes

Talked through a few cases verbally: the 3-island example, a grid that's all water, a single cell of land, and a checkerboard pattern to stress-test the diagonal rule.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and the expected output for the 3-island example, then outline a test strategy covering edge cases, typical cases, and the given example. Walk through each test case, explaining the input, expected output, and why it's nontrivial, and describe how you would verify the solution's correctness.

Pro tip: Demonstrate that you think about test coverage and edge cases beyond the provided example, such as empty input, single island, and maximum constraints, to show thoroughness and real-world engineering mindset.

1. Clarify the problem and expected behavior

Restate the problem to ensure you understand what constitutes a valid solution and what the output should represent for each test case.

2. Identify key scenarios and edge cases

List categories of test inputs: the given 3-island example, minimal cases (e.g., no islands, one island), maximal cases (e.g., all land, large grid), and cases with tricky connectivity (e.g., diagonal connections, holes).

3. Design specific test cases with expected outputs

For each scenario, define a concrete input and the correct output, ensuring the 3-island example is included and explained step-by-step.

4. Explain how to validate the solution

Describe how you would run the tests, compare outputs, and handle any discrepancies, possibly mentioning automated testing or manual walkthroughs.

5. Discuss test coverage and limitations

Summarize what the tests cover and acknowledge any untested scenarios, showing awareness of testing trade-offs.

Key Points to Mention

  • The 3-island example from the problem statement, with a clear walkthrough of the expected output.
  • Edge cases: empty grid, single cell, all water, all land, and grids with multiple disconnected islands.
  • Nontrivial inputs: grids with complex shapes, diagonal connections (if applicable), and large grids to test performance.
  • Correctness verification: comparing against a brute-force solution for small inputs or using known results.
  • Test automation: using unit tests or a test harness to run multiple cases efficiently.
  • Time and space complexity considerations when designing tests for large inputs.

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

Q5

How would you modify your solution if diagonal adjacency (8-directional) also counted as connected?

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Easy change in theory, just expand the directions array from 4 entries to 8.

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AI HintsAI Generated

Suggested Approach

First, clarify the original problem and solution (likely a grid connectivity problem using BFS/DFS with 4-directional moves). Then, explain that the core algorithm remains the same, but the neighbor exploration step expands from 4 to 8 directions. Finally, discuss any implications such as increased branching factor, potential performance impact, and edge cases like diagonal moves crossing obstacles.

Pro tip: Mention that diagonal moves can 'squeeze' through corners, which may or may not be allowed depending on the problem constraints—this shows attention to detail and real-world applicability.

1. Restate the original problem and solution

Briefly summarize the original problem (e.g., finding connected components in a grid) and your initial approach (e.g., BFS/DFS with 4-directional moves). This sets the context for the modification.

2. Identify the change: neighbor definition

Explain that the only change is in the definition of neighbors: from 4 (up, down, left, right) to 8 (including diagonals). The core algorithm (BFS/DFS) remains unchanged.

3. Adjust the neighbor exploration logic

Describe how to modify the code: add the four diagonal directions to the direction arrays or loops. Ensure boundary checks and visited marking still apply.

4. Analyze trade-offs and edge cases

Discuss performance implications (branching factor increases from 4 to 8, but still O(N) for N cells) and edge cases like diagonal moves crossing obstacles (if obstacles exist) or wrapping around boundaries.

5. Test and validate

Mention the importance of testing with cases that distinguish 4- vs 8-connectivity, such as a checkerboard pattern where diagonals connect otherwise separate components.

Key Points to Mention

  • Core algorithm (BFS/DFS) remains the same; only neighbor definition changes.
  • Direction arrays or loops need to include diagonals: (-1,-1), (-1,1), (1,-1), (1,1).
  • Time complexity remains O(N) for N cells, but constant factor increases due to more neighbors.
  • Edge case: diagonal moves may 'cut corners' through obstacles—clarify if allowed.
  • Visited marking must still prevent revisiting cells.
  • Testing should include cases where 8-connectivity merges components that 4-connectivity does not.

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