Problem Description

This is an interactive problem where you need to guide a robot through a hidden grid to find the shortest path from its starting position to a target cell.

The grid has dimensions m x n, but you don't know:

• The grid dimensions
• The robot's starting position
• The target cell's location
• Which cells are blocked or empty

You can only interact with the grid through a GridMaster object that provides three methods:

1. canMove(direction) - Returns true if the robot can move in the specified direction ('U', 'D', 'L', or 'R' for up, down, left, right), false if the move would hit a blocked cell or go off the grid
2. move(direction) - Moves the robot in the specified direction. If the move is invalid, the robot stays in place
3. isTarget() - Returns true if the robot is currently at the target cell

Your task is to determine the minimum distance (number of moves) from the robot's starting position to the target cell. If no valid path exists, return -1.

For testing purposes, the grid is represented as:

• -1 indicates the robot's starting position
• 0 indicates a blocked cell
• 1 indicates an empty cell
• 2 indicates the target cell

The grid is guaranteed to have exactly one starting position and one target, both of which are not blocked.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem involves navigating through a grid where the robot can move between cells (nodes) via directional movements (edges). The grid structure represents a graph where each cell is a node connected to its adjacent cells.

Is it a tree?

• No: The grid structure is not a tree. It can have cycles (the robot can return to previously visited cells), and there's no hierarchical parent-child relationship between cells.

Is the problem related to directed acyclic graphs?

• No: The grid allows bidirectional movement (if you can move from cell A to B, you can move back from B to A), and it can contain cycles.

Is the problem related to shortest paths?

• Yes: The problem explicitly asks for the "minimum distance" from the starting position to the target cell, which is a shortest path problem.

Is the graph weighted?

• No: Each move between adjacent cells has the same cost (1 step). All edges in this graph have uniform weight.

BFS

• Following the "no" branch from weighted graphs leads us to BFS for finding shortest paths in unweighted graphs.

Conclusion: While the flowchart suggests BFS for the shortest path component, the solution actually uses a two-phase approach:

1. DFS first to explore and map the reachable cells in the hidden grid (since we need to discover what cells exist and are accessible)
2. BFS then to find the shortest path from start to target using the mapped information

The DFS pattern is essential here because we need to exhaustively explore the unknown grid structure before we can compute shortest paths.

Intuition

The key challenge in this problem is that we're working with a hidden grid - we can't see the entire map at once. It's like being blindfolded in a maze and only being able to feel the walls around you. This fundamentally changes how we approach the shortest path problem.

In a typical shortest path problem where we can see the entire grid, we'd simply run BFS from the start to find the target. But here, we first need to discover what the grid looks like. Think of it as a two-phase exploration:

Phase 1: Map Discovery We need to explore the entire reachable area to understand the grid's structure. DFS is perfect for this exhaustive exploration because:

• We can systematically visit every reachable cell
• After moving in one direction, we can backtrack to our previous position (using the opposite direction)
• We can mark the target location when we find it
• We build a set of all valid, reachable positions

The backtracking aspect is crucial - when we move('U'), we can return to our original position with move('D'). This allows us to fully explore all branches from each position, essentially mapping out the accessible parts of the grid.

Phase 2: Finding the Shortest Path Once we've mapped the grid and know where the target is, we have transformed the problem into a standard shortest path problem on a known graph. Now we can use BFS on our discovered map to find the minimum distance from start (0, 0) to the target coordinates.

The elegance of this solution is in recognizing that we need to separate the exploration problem from the pathfinding problem. We can't find the shortest path to a destination we haven't discovered yet, so we first use DFS's exhaustive nature to uncover the map, then use BFS's level-by-level expansion to find the optimal path.

Learn more about Depth-First Search and Breadth-First Search patterns.

Solution Approach

The implementation follows the two-phase approach we identified: DFS for exploration and BFS for finding the shortest path.

Phase 1: DFS Exploration and Grid Mapping

We start by assuming the robot begins at coordinate (0, 0). The DFS function explores the grid recursively:

def dfs(i: int, j: int):
    if master.isTarget():
        nonlocal target
        target = (i, j)  # Record target coordinates
        return

For each position, we try moving in all four directions using the pattern s = "URDL" (Up, Right, Down, Left). The direction offsets are stored in dirs = (-1, 0, 1, 0, -1) where consecutive pairs represent (dx, dy) for each direction.

The key insight is the backtracking mechanism:

for k, c in enumerate(s):
    x, y = i + dirs[k], j + dirs[k + 1]
    if master.canMove(c) and (x, y) not in vis:
        vis.add((x, y))
        master.move(c)
        dfs(x, y)
        master.move(s[(k + 2) % 4])  # Backtrack using opposite direction

The opposite direction is calculated using (k + 2) % 4:

• Up (0) ↔ Down (2)
• Right (1) ↔ Left (3)

This ensures the robot returns to its previous position after exploring each branch, allowing complete exploration of the reachable grid.

Phase 2: BFS for Shortest Path

After exploration, if no target was found (target is None), we return -1. Otherwise, we perform BFS on the discovered map:

vis.discard((0, 0))  # Remove start from visited set
q = deque([(0, 0)])
ans = -1

The BFS proceeds level by level, incrementing the distance counter for each level:

while q:
    ans += 1
    for _ in range(len(q)):
        i, j = q.popleft()
        if (i, j) == target:
            return ans
        for a, b in pairwise(dirs):
            x, y = i + a, j + b
            if (x, y) in vis:
                vis.remove((x, y))
                q.append((x, y))

The pairwise(dirs) function generates consecutive pairs from the direction array, giving us the (dx, dy) offsets for each direction. We use the vis set as our "unvisited" set - when we visit a cell during BFS, we remove it from vis to avoid revisiting.

Data Structures Used:

• vis: A set storing all reachable coordinates discovered during DFS
• target: A tuple storing the target cell's coordinates
• q: A deque for BFS queue operations
• dirs and s: Arrays for direction management

The time complexity is O(m × n) for both DFS exploration and BFS pathfinding, where m and n are the grid dimensions. The space complexity is also O(m × n) for storing the visited cells.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Consider this hidden grid:

Grid (hidden from us):
[-1][ 1][ 0]
[ 1][ 1][ 1]
[ 0][ 1][ 2]

-1 = start, 0 = blocked, 1 = empty, 2 = target

The robot starts at the top-left corner, which we'll call position (0, 0) in our coordinate system.

Phase 1: DFS Exploration

Starting at (0, 0), we begin exploring:

1. At (0, 0): Check all directions using "URDL" order
   • Up: canMove('U') returns false (off grid)
   • Right: canMove('R') returns true → Move to (0, 1)
2. At (0, 1): Continue exploring
   • Up: false (off grid)
   • Right: false (blocked cell)
   • Down: true → Move to (1, 1)
3. At (1, 1): The center of reachable area
   • Up: already visited (0, 1)
   • Right: true → Move to (1, 2)
4. At (1, 2):
   • Check isTarget() → false
   • Down: true → Move to (2, 2)
5. At (2, 2):
   • Check isTarget() → true! Record target = (2, 2)
   • Backtrack to (1, 2) using move('U')
6. Continue backtracking and exploring unvisited branches:
   • From (1, 1): Try left → Move to (1, 0)
   • From (1, 0): No new paths, backtrack

After complete exploration, our vis set contains: {(0, 0), (0, 1), (1, 0), (1, 1), (1, 2), (2, 2)}

Our discovered map looks like:

[S][✓][ ]
[✓][✓][✓]
[ ][✓][T]

S = start, ✓ = reachable, T = target

Phase 2: BFS for Shortest Path

Now we find the shortest path from (0, 0) to target (2, 2):

1. Level 0 (distance = 0): Start with queue = [(0, 0)]

   • Process (0, 0): Not target, add neighbors
   • Can go right to (0, 1) and down to (1, 0)
   • Queue becomes: [(0, 1), (1, 0)]

2. Level 1 (distance = 1): Process 2 cells

   • Process (0, 1): Not target, add (1, 1)
   • Process (1, 0): Not target, add (1, 1) (already in queue)
   • Queue becomes: [(1, 1)]

3. Level 2 (distance = 2): Process 1 cell

   • Process (1, 1): Not target, add (1, 2)
   • Queue becomes: [(1, 2)]

4. Level 3 (distance = 3): Process 1 cell

   • Process (1, 2): Not target, add (2, 2)
   • Queue becomes: [(2, 2)]

5. Level 4 (distance = 4): Process 1 cell

   • Process (2, 2): Target found! Return distance = 4

The shortest path is 4 moves: (0,0) → (0,1) → (1,1) → (1,2) → (2,2)

This example demonstrates how DFS first maps the entire reachable grid through backtracking exploration, then BFS efficiently finds the shortest path using the discovered map.

Solution Implementation

Time and Space Complexity

Time Complexity: O(N + N) = O(N), where N is the number of accessible cells in the grid.

• The DFS exploration phase visits each accessible cell exactly once. For each cell, it checks 4 directions and potentially makes moves/backtracks, resulting in O(N) time for the DFS traversal.
• The BFS phase for finding the shortest path also visits each accessible cell at most once. Each cell is added to the queue once and processed once, with constant-time operations for checking 4 directions. This also takes O(N) time.
• Overall time complexity: O(N) + O(N) = O(N)

Space Complexity: O(N), where N is the number of accessible cells in the grid.

• The vis set stores all accessible cells discovered during DFS, which requires O(N) space.
• The BFS queue can contain at most O(N) cells in the worst case (though typically much less).
• The recursive call stack for DFS can go as deep as O(N) in the worst case (e.g., a long snake-like path).
• Since these space usages don't accumulate (the recursion stack is released before BFS starts), the overall space complexity is O(N).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Backtracking Direction Calculation

One of the most common mistakes is incorrectly calculating the opposite direction for backtracking after DFS exploration. Many developers might try:

• Using (k + 1) % 4 or (k + 3) % 4 instead of (k + 2) % 4
• Hardcoding opposite directions with if-else statements that may have errors
• Forgetting that the direction order matters (URDL vs UDLR changes the calculation)

Why this happens: The modulo arithmetic for finding opposite directions isn't intuitive. With "URDL" ordering:

• Up (index 0) opposite is Down (index 2)
• Right (index 1) opposite is Left (index 3)
• Down (index 2) opposite is Up (index 0)
• Left (index 3) opposite is Right (index 1)

Solution: Always verify your direction mapping with a simple test or use a dictionary for clarity:

OPPOSITE = {'U': 'D', 'D': 'U', 'L': 'R', 'R': 'L'}
# Then use: master.move(OPPOSITE[direction_char])

2. Not Properly Managing the Visited Set Between DFS and BFS

A critical pitfall is mishandling the visited_cells set when transitioning from DFS to BFS:

• Forgetting to remove the starting position (0, 0) from the visited set before BFS
• Using the same visited tracking mechanism for both phases without resetting
• Adding (0, 0) to visited during DFS exploration

Why this happens: During DFS, we mark cells as visited to avoid cycles. However, for BFS, we need to traverse these cells again to find the shortest path. The starting position (0, 0) is never explicitly added to visited_cells during DFS but needs special handling.

Solution: Use visited_cells.discard((0, 0)) instead of remove() to safely handle the case where (0, 0) might not be in the set, avoiding a KeyError.

3. Forgetting to Return Robot to Original Position After DFS

If the robot doesn't return to (0, 0) after DFS exploration completes, the BFS phase will calculate distances from the wrong starting position.

Why this happens: The recursive DFS with backtracking should naturally return the robot to start, but any error in the backtracking logic or early returns can leave the robot stranded.

Solution: Consider adding a verification step or explicitly tracking the robot's position:

# After DFS completes, verify robot is at start
# Could add debug check: assert current_position == (0, 0)

4. Misaligning Direction Characters with Delta Arrays

The indices in DIRECTIONS = "URDL" must perfectly align with the delta pairs in DIRECTION_DELTAS = (-1, 0, 1, 0, -1). A common mistake is:

• Changing the order of directions without updating deltas
• Using wrong index calculations when accessing deltas
• Mixing up row/column order in deltas

Why this happens: The compact representation (-1, 0, 1, 0, -1) where consecutive pairs represent (dx, dy) is efficient but error-prone.

Solution: Use a more explicit data structure:

MOVES = {
    'U': (-1, 0),
    'R': (0, 1),
    'D': (1, 0),
    'L': (0, -1)
}
# Then iterate: for direction, (dx, dy) in MOVES.items():

5. Off-by-One Error in BFS Distance Calculation

Starting distance = -1 and incrementing at the beginning of each BFS level is correct, but developers often mistakenly:

• Start with distance = 0 and increment after processing each level
• Return distance + 1 when finding the target
• Increment distance inside the inner loop instead of the outer loop

Why this happens: The pattern of incrementing before processing each level (starting at -1) is less common than incrementing after.

Solution: Think of it as: "We're about to process level K, so increment the distance counter to K." The first level (distance 0) contains only the starting position.