The infinite grid part is what tripped me up initially.
Use BFS with a queue to explore the grid level by level, marking visited cells to avoid cycles. Since the grid is infinite, bound the search by the Manhattan distance from start to goal; if the goal is unreachable, BFS will exhaust all reachable cells within that bound and terminate.
Pro tip: Mention that you can optimize by using a bidirectional BFS or A* with Manhattan distance heuristic, but for an infinite grid, the key is to set a termination condition based on the maximum possible distance to the goal.
Confirm that the grid is infinite, movements are 4-directional, and blocked cells are static. Discuss how to represent the grid (e.g., hash set for blocked cells) and visited set.
Explain that BFS guarantees shortest path in unweighted graphs. For infinite grid, set a bound: if the queue is empty or the current distance exceeds Manhattan distance from start to goal, terminate and return no path.
Use a queue for BFS, a set for visited coordinates, and a parent map to reconstruct the path. For each cell, explore neighbors, skip blocked or visited cells, and enqueue valid ones.
Once the goal is reached, backtrack using the parent map to build the path from start to goal. If the queue empties without reaching the goal, return that no path exists.
Discuss time and space complexity in terms of the number of reachable cells within the bound. Mention edge cases: start equals goal, start or goal blocked, and unreachable goal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Model the problem as a time-expanded graph where each state is (position, time mod 2) because obstacles shift every 2 seconds. Use BFS to find the shortest path in this graph, checking at each step whether the next position is safe at the next time. If the target is reached, return the time; if BFS exhausts all reachable states, return unreachable.
Pro tip: Mention that the state space is finite modulo the obstacle period, so BFS will terminate; also discuss how to handle large time values by noting that if a path exists, it will be found within a bounded number of steps due to the periodic nature.
Clarify that obstacles move periodically every 2 seconds, and you can move or wait each second. The grid is infinite, so you need an efficient search that doesn't explore infinitely.
Represent each state as (x, y, t mod 2) because the obstacle configuration repeats every 2 seconds. This reduces the infinite time dimension to a finite set of states per position.
From each state, consider moving to adjacent cells or waiting. For each action, compute the next time and check if the destination cell is free of obstacles at that time. Use BFS to find the shortest path in terms of time.
Since the state space is finite (positions modulo obstacle period), BFS will either find the target or exhaust all reachable states. If exhausted, return unreachable.
Discuss time and space complexity: O(V+E) where V is the number of reachable states. Consider optimizations like bidirectional BFS or A* if applicable, and note that the infinite grid is effectively bounded by the reachable area.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.