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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Bytedance SWE interview, coding round. One algorithm question, pretty standard grid traversal stuff. Nothing unexpected but the pressure is real.

Questions Asked (1)

Q1

Given a 2D grid with open cells and walls, find the shortest path from a start cell to an end cell. Return the path length or -1 if no path exists.

Algorithms & Data Structures
Author's notes

Classic BFS, four directions, nothing tricky.

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AI HintsAI Generated

Suggested Approach

Model the grid as an unweighted graph and use BFS to find the shortest path from start to end. Track visited cells to avoid cycles and return the distance when the end is reached, or -1 if the queue is exhausted.

Pro tip: Mention that BFS is optimal for unweighted grids, but if the grid were weighted (e.g., different terrain costs), you'd switch to Dijkstra's algorithm. Also, discuss early termination when the end is found to save time.

1. Clarify the problem

Confirm the grid dimensions, movement allowed (4-directional or 8-directional), and whether diagonal moves are permitted. Ask if the start and end are guaranteed to be open cells.

2. Choose BFS

Explain that BFS is ideal because it explores level by level, guaranteeing the shortest path in an unweighted graph. Mention that DFS would not guarantee the shortest path.

3. Outline BFS algorithm

Initialize a queue with the start cell and a visited set. While the queue is not empty, dequeue a cell, check if it's the end, and enqueue all valid unvisited neighbors (within bounds, not walls). Track distance by storing (cell, distance) or using level-order traversal.

4. Handle edge cases

Consider cases where start equals end (return 0), start or end is a wall (return -1), or the grid is empty. Also, discuss memory optimization for large grids (e.g., using a 2D boolean array for visited).

5. Analyze complexity

State that time complexity is O(R*C) where R and C are grid dimensions, as each cell is visited at most once. Space complexity is O(R*C) for the queue and visited set in the worst case.

Key Points to Mention

  • BFS guarantees shortest path in unweighted graphs
  • Use a queue for BFS and a visited set/array to avoid revisiting cells
  • Check boundaries and wall conditions before enqueueing neighbors
  • Early termination when the end cell is dequeued
  • Time and space complexity: O(R*C)
  • Alternative: Dijkstra's algorithm if the grid has weighted cells

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