Standard BFS with a visited set, nothing too tricky.
Model the grid as an unweighted graph and use BFS from the top-left cell to find the shortest path to the bottom-right, tracking distance as you go. Clarify edge cases (start/end blocked, empty grid) and state time/space complexity before coding.
Pro tip: Mention that BFS guarantees the shortest path in an unweighted grid, and note that you can optimize space by marking visited cells in-place or using a direction array—this shows attention to practical constraints.
Confirm grid dimensions, movement allowed (4-directional), and handle cases like start/end blocked, 1x1 grid, or no path. State assumptions clearly.
Explain that BFS explores level by level, guaranteeing the shortest path in an unweighted grid, unlike DFS which may find a longer path first.
Describe using a queue storing (row, col, distance), a visited set or in-place marking, and checking bounds and blocked cells before enqueuing neighbors.
State that time complexity is O(m*n) since each cell is visited at most once, and space complexity is O(m*n) for the queue and visited structure in the worst case.
Mention bidirectional BFS for potential speedup, or using a distance matrix if the grid must remain unchanged. Briefly note how to adapt for 8-directional movement.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
My first instinct was to iterate over every blocked cell, flip it, run BFS each time, and take the minimum.
Model the problem as a shortest path on an expanded state graph where each state includes the number of blocked cells converted so far (0 or 1). Run BFS from the start, allowing transitions through free cells with the same conversion count and through blocked cells only if the count is 0 (incrementing it to 1). The first time you reach the destination with either conversion count gives the shortest path length.
Pro tip: Mention that the expanded graph has at most 2*N*M states, so BFS remains O(N*M) and is optimal for unweighted grids; this shows you understand both correctness and complexity trade-offs.
Represent each state as (row, col, used_conversion) where used_conversion is 0 or 1. This captures whether the single allowed conversion has been used.
From a state, move to adjacent free cells with the same used_conversion; move to adjacent blocked cells only if used_conversion is 0, transitioning to used_conversion=1.
Start BFS from (start_row, start_col, 0) with distance 0. Use a queue and a visited array of size N x M x 2 to avoid revisiting states.
When you dequeue the destination cell with either used_conversion value, return the current distance as the shortest path length. If BFS exhausts without reaching it, return -1.
State the time and space complexity: O(N*M) since each cell is visited at most twice (once per conversion count), and the queue/visited structures are proportional to 2*N*M.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.