Minimum wall breaks
You are given an R x C grid where each cell is 0 (open) or 1 (a wall). Starting at the top-left (0,0) you may move up/down/left/right; entering an open cell is free but entering a wall costs 1 'break'. Return the minimum number of walls you must break to reach the bottom-right cell (R-1, C-1). The start and goal cells contribute their own cell cost if they are walls. Assume 1 <= R, C <= 1000, so a 0-1 BFS (O(R*C)) is expected rather than Dijkstra with a heap.
Implement
min_wall_breaks(grid: list[list[int]]) → intExamples
in
[[[0,1,0],[0,1,0],[0,0,0]]]out0What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 25 min
solution.py
InputExpectedGot
[[[0,1,0],[0,1,0],[0,0,0]]]0not run yetsample