Shortest Path on a Grid with Alternating Move Directions
You have an $n \times n$ grid where some cells are obstacles. Movement rules alternate by time step:
- At even time steps ($t = 0, 2, 4, \ldots$), you may only move vertically (up or down by one cell).
- At odd time steps ($t = 1, 3, 5, \ldots$), you may only move horizontally (left or right by one cell).
You may also choose to stay in place at any time step (this still advances the clock by 1).
You start at cell $(1, 1)$ at time $t = 0$ and want to reach cell $(n, n)$.
- Define a state space that accounts for position and time parity. How large is this state space?
- Design an $O(n^2)$ algorithm to compute the minimum time to reach $(n, n)$, or report that it is unreachable.
- Prove that BFS on your state space returns the correct shortest-path answer.
Constraints: - $2 \leq n \leq 1000$ - The grid is given as an $n \times n$ binary matrix where $0$ = free cell and $1$ = obstacle. - $(1,1)$ and $(n,n)$ are always free.
Example 1: ``` Input: grid = [[0,0,0], [0,1,0], [0,0,0]] Output: 6 Explanation (0-indexed cells; each move advances the clock by 1): t=0 (even, vertical): (0,0) -> (1,0) t=1 (odd, horizontal): (1,0) -> (1,0) [stay; obstacle at (1,1)] t=2 (even, vertical): (1,0) -> (2,0) t=3 (odd, horizontal): (2,0) -> (2,1) t=4 (even, vertical): (2,1) -> (2,1) [stay] t=5 (odd, horizontal): (2,1) -> (2,2) The target (2,2) is first occupied after 6 moves, so the minimum time is 6. ```
Example 2: ``` Input: grid = [[0,0], [0,0]] Output: 2 Explanation (0-indexed): t=0 (even, vertical): (0,0) -> (1,0) t=1 (odd, horizontal): (1,0) -> (1,1) (1,1) is (n,n) for n=2, reached after 2 moves. Minimum time = 2. ```
Hints
- Standard BFS on just the grid coordinates is not enough -- what extra information do you need to track to know which moves are currently allowed?
- Think of the grid as two layers: one for even time steps (vertical moves) and one for odd time steps (horizontal moves). Every move crosses from one layer to the other.
- Define states as $(r, c, p)$ where $p \in \{0, 1\}$ is the time parity. The state space has $2n^2$ nodes, so BFS runs in $O(n^2)$. Don't forget to allow staying in place as a valid transition.
Worked Solution
How to Think About It: The twist here is that the allowed move direction depends on whether the current time step is even or odd. A standard BFS on grid coordinates alone loses track of which directions are available right now. The fix is simple: expand the state to include the time parity. Since parity only has two values (even or odd), this only doubles the state space -- still $O(n^2)$.
Think of it as two copies of the grid layered on top of each other: the "even layer" (where you can move vertically) and the "odd layer" (where you can move horizontally). Every transition moves you from one layer to the other. BFS on this layered graph gives shortest paths because all edges have weight 1.
Algorithm:
- State space. Define $S = \{(r, c, p) : 1 \leq r, c \leq n, \; p \in \{0, 1\}\}$ where $p$ is the time parity ($0$ = even, $1$ = odd). The size of $S$ is $2n^2$.
2. Transitions from state $(r, c, p)$: - If $p = 0$ (even step -- vertical moves allowed): neighbors are $(r-1, c, 1)$ and $(r+1, c, 1)$, plus $(r, c, 1)$ for staying in place. - If $p = 1$ (odd step -- horizontal moves allowed): neighbors are $(r, c-1, 0)$ and $(r, c+1, 0)$, plus $(r, c, 0)$ for staying in place. - A neighbor is valid only if it is within bounds and the target cell is not an obstacle.
- BFS. Initialize a queue with the start state $(1, 1, 0)$ at distance $0$. Run standard BFS. The answer is $\min(\text{dist}(n, n, 0), \text{dist}(n, n, 1))$ -- we accept reaching $(n, n)$ on either parity.
Code:
```python from collections import deque
def min_time_to_reach(grid): n = len(grid) # dist[r][c][p] = minimum time to reach (r,c) with parity p INF = float('inf') dist = [[[INF, INF] for _ in range(n)] for _ in range(n)] dist[0][0][0] = 0 # start at (0,0), parity 0, time 0
queue = deque() queue.append((0, 0, 0)) # (row, col, parity)
while queue: r, c, p = queue.popleft() t = dist[r][c][p] np = 1 - p # next parity
if p == 0: # Even step: vertical moves (up, down) or stay moves = [(r - 1, c), (r + 1, c), (r, c)] else: # Odd step: horizontal moves (left, right) or stay moves = [(r, c - 1), (r, c + 1), (r, c)]
for nr, nc in moves: if 0 <= nr < n and 0 <= nc < n and grid[nr][nc] == 0: if dist[nr][nc][np] > t + 1: dist[nr][nc][np] = t + 1 queue.append((nr, nc, np))
ans = min(dist[n - 1][n - 1][0], dist[n - 1][n - 1][1]) return ans if ans < INF else -1 ```
Correctness Proof:
We need to show BFS on the state graph $G = (S, E)$ returns the shortest path.
- Unweighted graph. Every edge in $G$ has weight 1 (each transition takes exactly one time step). BFS on an unweighted graph finds shortest paths -- this is a standard result.
- State space captures all information. At any point during traversal, the next set of allowed moves depends only on the current cell $(r, c)$ and the parity $p$ of the current time step. The state $(r, c, p)$ is a sufficient statistic for the future -- there is no hidden state. Therefore any path in the original problem corresponds to a path in $G$, and vice versa.
- Optimality. Suppose there is a path in the original problem reaching $(n, n)$ in $T$ steps. This path corresponds to a sequence of states in $G$ of length $T$. BFS finds the shortest such sequence. Conversely, any path in $G$ corresponds to a valid sequence of moves in the original problem. So the BFS distance equals the minimum time.
- Complexity. The state space has $|S| = 2n^2$ nodes. Each node has at most 3 outgoing edges. BFS visits each node at most once. Total work: $O(n^2)$ time and $O(n^2)$ space.
Answer: Define the state space $S = \{(r, c, p)\}$ with $|S| = 2n^2$. Run BFS from $(1, 1, 0)$ on the layered graph where even-parity states connect vertically and odd-parity states connect horizontally. BFS returns the shortest path in $O(n^2)$ time because the graph is unweighted and the state fully determines available transitions.
Intuition
The core technique here is state-space augmentation: when the rules of movement depend on some extra variable (here, the parity of the time step), you fold that variable into the state. Instead of searching on the raw grid, you search on a graph that has multiple "copies" of each cell -- one per value of the extra variable. This is the same idea behind modeling problems with day/night cycles, alternating turn games, or any system where the transition rules are periodic. The state space grows by a factor equal to the period (here, 2), which is cheap.
This pattern shows up constantly in quant and CS interviews. Shortest-path problems on grids with constraints (fuel, keys, time-dependent edges) almost always reduce to BFS or Dijkstra on an augmented state space. The key insight to communicate in an interview is: identify the minimal extra state needed to make transitions memoryless (Markovian), augment the graph, and then apply the standard shortest-path algorithm. If all edge weights are 1, BFS suffices. If weights vary, use Dijkstra. The augmented graph is always small as long as the extra variable has a small domain.