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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
May 2026

Summary

Bytedance SWE interview with a grid pathfinding problem that sounds manageable until you realize the state space is three-dimensional. The k-obstacle-removal constraint is what makes it a real question rather than just BFS.

Questions Asked (1)

Q1

Given an m x n grid of empty cells and obstacles, find the shortest path from the top-left to the bottom-right corner where you can remove at most k obstacles along the way. Return the number of steps, or -1 if no path exists.

Algorithms & Data Structures
Author's notes

My first instinct was plain BFS and I started coding it before catching myself.

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AI HintsAI Generated

Suggested Approach

Model the problem as a shortest path on a state graph where each state is (row, col, obstacles_removed). Use BFS since each move costs 1 step, and track the minimum obstacles removed to reach each cell to prune dominated states. Return the distance when reaching the bottom-right with obstacles_removed ≤ k, else -1.

Pro tip: Emphasize that BFS guarantees the shortest path in unweighted graphs, and mention that you can optimize by storing the minimum obstacles removed per cell to avoid revisiting with worse states. This shows you understand both correctness and efficiency.

1. Clarify the problem and constraints

Confirm grid dimensions, movement directions (usually 4-directional), and that removing an obstacle counts toward k. Ask about edge cases like start or end being obstacles.

2. Define the state and transition

State: (r, c, removed). From (r, c), move to adjacent cells; if the neighbor is an obstacle, increment removed by 1. Only allow moves where removed ≤ k.

3. Choose BFS with state tracking

Use a queue for BFS. Maintain a 2D array minRemoved[r][c] to store the minimum obstacles removed to reach (r, c). Only enqueue a state if it improves minRemoved[r][c].

4. Handle termination and return result

When popping a state at (m-1, n-1), return the current step count. If the queue empties without reaching the target, return -1.

5. Analyze complexity and edge cases

Time: O(m*n*k) worst-case, but with pruning often O(m*n). Space: O(m*n). Discuss edge cases: k=0, start/end blocked, no path.

Key Points to Mention

  • BFS is optimal for unweighted shortest path problems.
  • State space includes obstacles removed, not just position.
  • Pruning via minRemoved array avoids redundant work.
  • Time and space complexity analysis.
  • Edge cases: start or end is an obstacle, k=0, unreachable target.
  • Alternative approaches like Dijkstra with 0-1 BFS or A* if weights vary.

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