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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Apr 2026

Summary

Bytedance technical phone screen, got hit with the classic number of islands problem and had to walk through both DFS and BFS live. Pretty standard for this kind of role but the complexity discussion at the end tripped me up more than I expected.

Questions Asked (1)

Q1

Given a 2D binary grid where '1' is land and '0' is water, count the number of islands, where an island is a group of '1's connected horizontally or vertically. Walk through both DFS and BFS solutions and their time/space complexity.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

I went DFS first because it felt more natural to code quickly, just mark visited cells as '0' and recurse in four directions.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem constraints and edge cases, then explain the core idea of traversing each unvisited land cell and marking all connected land cells as visited. Walk through both DFS and BFS implementations, highlighting their trade-offs, and conclude with time and space complexity analysis.

Pro tip: Mention that DFS can cause stack overflow for large grids, so BFS or iterative DFS is safer in production; also note that modifying the input grid in-place saves space but may not be allowed if the input must be preserved.

1. Clarify and Restate

Confirm the grid dimensions, connectivity definition (4-directional), and whether the input can be modified. Restate the problem to ensure alignment with the interviewer.

2. High-Level Strategy

Explain that you'll iterate through each cell; when you find a '1', increment the island count and traverse all connected '1's using DFS or BFS, marking them as visited (e.g., set to '0').

3. DFS Walkthrough

Describe the recursive DFS approach: from a starting cell, recursively visit all adjacent land cells in four directions, marking them as visited. Mention base cases and recursion depth concerns.

4. BFS Walkthrough

Describe the iterative BFS approach: use a queue to explore all adjacent land cells level by level, marking them as visited upon enqueue. Highlight that BFS avoids recursion depth issues.

5. Complexity and Trade-offs

State that both DFS and BFS have O(M×N) time and O(M×N) space in the worst case (due to recursion stack or queue). Discuss trade-offs: DFS is simpler but risks stack overflow; BFS uses more memory but is safer for large grids.

Key Points to Mention

  • Time complexity: O(M×N) where M is rows and N is columns, as each cell is visited once.
  • Space complexity: O(M×N) worst case for both DFS (recursion stack) and BFS (queue), e.g., all land cells.
  • In-place modification: marking visited cells by setting them to '0' avoids extra space but alters input.
  • DFS vs BFS trade-offs: DFS may cause stack overflow for large grids; BFS uses explicit queue and is iterative.
  • Edge cases: empty grid, single row/column, all water, all land.
  • Alternative: Union-Find (Disjoint Set) can also solve this, but may be overkill; mention if asked for alternatives.

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