The fix itself isn't hard once you see it.
First, explain that BFS guarantees the shortest path in an unweighted grid, but only if each cell is enqueued at most once. Then, describe adding a visited set or 2D boolean array, marking cells as visited when enqueued, and returning the distance when the end cell is dequeued (or -1 if the queue empties).
Pro tip: Mention that marking visited at enqueue time (not dequeue) prevents duplicate entries and is the standard BFS pattern; also note that you can mutate the grid in-place to save memory if allowed.
Confirm the grid dimensions, movement directions (4-way or 8-way), and whether obstacles are represented as blocked cells. Ask if modifying the input grid is acceptable.
Explain that without tracking visited cells, the same cell can be enqueued multiple times, leading to exponential time and potential infinite loops. This breaks BFS's O(V+E) complexity.
Add a visited data structure (e.g., a 2D boolean array or a set of coordinates). Mark a cell as visited immediately when it is enqueued, not when dequeued, to avoid duplicates.
Use a queue storing (row, col, distance) or process level by level. Return the distance when the end cell is reached; if the queue empties, return -1.
State that time and space are O(rows * cols). Test cases: start equals end, unreachable end, empty grid, and single row/column.
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