I went down a BFS path for a few minutes before realizing that was completely overkill.
Model the problem as a reachability question: at each time step, the second vehicle can reach any cell within a Manhattan distance equal to the time elapsed. Check if the truck's position at each time step is within that reachable set. If at any time step the Manhattan distance from the origin to the truck's position is ≤ the time step, then the second vehicle can intercept the truck.
Pro tip: Clarify that the second vehicle can wait, so reachability is not just about shortest paths but about parity and timing. Mention that if the truck's position at time t satisfies |x| + |y| ≤ t and the parity of (|x| + |y|) matches t, then interception is possible; otherwise, it might still be possible if the vehicle can wait, but parity matters for exact timing.
Restate the problem: given the truck's positions at each time step, determine if the second vehicle starting at (0,0) can be at the same cell at the same time. The second vehicle moves one step in cardinal directions or stays put each turn.
At time t, the second vehicle can reach any cell (x,y) such that |x| + |y| ≤ t and (|x| + |y|) % 2 == t % 2. This is because each move changes the Manhattan distance from the origin by at most 1, and waiting preserves parity.
Iterate through the truck's positions at each time step t. For each position (x,y), check if it is reachable by the second vehicle at time t using the condition above. If any is reachable, return true.
If no time step allows interception, return false. Otherwise, return true as soon as a reachable position is found.
The algorithm runs in O(n) time where n is the number of time steps, and O(1) extra space. This is optimal since we must examine each time step.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.