Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Bytedance coding screen, one question the whole time. They threw a constraint I wasn't expecting and it changed everything about how I approached it.

Questions Asked (1)

Given an m×n grid of '1's (land) and '0's (water), count the number of islands. You must use BFS and cannot modify the grid in place; track visited cells with a separate data structure.

Algorithms & Data Structures

Author's notes

I've done number of islands before so I felt fine at first, then they said no in-place modification.

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AI HintsAI Generated

Suggested Approach

Use BFS to explore each unvisited land cell, marking all connected land cells as visited in a separate 2D boolean array. Iterate through every cell; when an unvisited '1' is found, increment the island count and launch BFS to mark its entire island.

Pro tip: Clarify upfront that you're using a visited matrix to avoid mutating the input, and mention that BFS is preferred over DFS here to avoid recursion depth issues on large grids.

1. Clarify constraints and edge cases

Confirm grid dimensions, whether diagonal adjacency counts (usually no), and that the grid must remain unmodified. Discuss handling of empty grids or grids with no land.

2. Initialize visited structure and island counter

Create a 2D boolean array of the same size as the grid, initially all false, and set island count to 0.

3. Iterate through each cell

For each cell (i, j), if it's land ('1') and not visited, increment the island count and start a BFS from that cell.

4. Perform BFS to mark the island

Use a queue to explore all connected land cells (up, down, left, right). For each neighbor that is land and unvisited, mark it visited and enqueue it.

5. Return the island count

After processing all cells, return the total number of islands found.

Key Points to Mention

Use a separate visited matrix to avoid modifying the input grid.
BFS traversal with a queue to explore all 4-directionally connected land cells.
Time complexity: O(m*n) since each cell is visited at most once.
Space complexity: O(m*n) for the visited matrix and queue in the worst case.
Edge cases: empty grid, all water, all land, single row/column.
Clarify that diagonal connections are not considered unless specified.

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