Solution Approach

The implementation uses DFS with a visited matrix to track which stopping positions we've already explored. Let's walk through the key components:

1. Visited Matrix Setup:

vis = [[False] * n for _ in range(m)]

We create a 2D boolean matrix matching the maze dimensions to track visited stopping positions. This prevents infinite loops when the ball revisits the same stopping point.

2. DFS Function Structure: The dfs(i, j) function explores all possible paths from position (i, j):

• First, it checks if this position was already visited. If yes, return immediately to avoid redundant exploration
• Mark the current position as visited: vis[i][j] = True
• Check if we've reached the destination: if [i, j] == destination

3. Rolling Simulation:

for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:  x, y = i, j  while 0 <= x + a < m and 0 <= y + b < n and maze[x + a][y + b] == 0:  x, y = x + a, y + b  dfs(x, y)

For each of the 4 directions (left, right, down, up):

• Initialize (x, y) to the current position
• Use a while loop to simulate the ball rolling in direction (a, b)
• The ball keeps rolling while:
  • The next position (x + a, y + b) is within bounds
  • The next position is an empty space (value is 0)
• When the loop exits, (x, y) is the stopping position
• Recursively call dfs(x, y) to explore from this new stopping position

4. Main Execution:

dfs(start[0], start[1]) return vis[destination[0]][destination[1]]

• Start the DFS from the initial position
• After DFS completes, check if the destination was visited
• If vis[destination[0]][destination[1]] is True, the ball can reach and stop at the destination

Why This Works: The algorithm systematically explores all reachable stopping positions from the start. By marking positions as visited, we ensure each stopping position is processed only once. If the destination becomes marked as visited during this exploration, it means there exists at least one valid path for the ball to reach and stop at the destination.

The time complexity is O(m*n*max(m,n)) where the extra factor accounts for the rolling simulation in each direction. The space complexity is O(m*n) for the visited matrix and recursion stack.

Example Walkthrough

Let's walk through a small example to illustrate how the DFS solution works.

Consider this maze:

maze = [  [0, 0, 1, 0, 0],  [0, 0, 0, 0, 0],  [0, 0, 0, 1, 0],  [1, 1, 0, 1, 1],  [0, 0, 0, 0, 0] ] start = [0, 4] destination = [4, 4]

Where 0 = empty space, 1 = wall, S = start position (0,4), D = destination (4,4).

Visualized with S and D:

[0, 0, 1, 0, S] [0, 0, 0, 0, 0] [0, 0, 0, 1, 0] [1, 1, 0, 1, 1] [0, 0, 0, 0, D]

Step 1: Start DFS from position (0, 4)

• Mark (0, 4) as visited
• Try rolling in 4 directions:

Step 2: Rolling Left from (0, 4)

• Ball rolls: (0,4) → (0,3) → stops at (0,2) because maze[0][2] = 1 (wall)
• Call dfs(0, 2)
• Mark (0, 2) as visited, but it's not the destination
• From (0, 2), the ball can only roll down

Step 3: Rolling Down from (0, 2)

• Ball rolls: (0,2) → (1,2) → (2,2) → stops at (2,2) because maze[3][2] = 0 but maze[3][1] = 1 and maze[3][3] = 1 (surrounded by walls)
• Call dfs(2, 2)
• Mark (2, 2) as visited

Step 4: From (2, 2), try all directions

• Up: rolls back to (0, 2) - already visited, skip
• Down: can't move (wall at 3,2)
• Left: rolls to (2, 0) - stops at left boundary
• Right: rolls to (2, 3) - stops at wall

Step 5: Continue exploring from (2, 0)

• Mark (2, 0) as visited
• Roll down: goes to (4, 0)
• From (4, 0), roll right: (4,0) → (4,1) → (4,2) → (4,3) → (4,4)
• Ball stops at (4, 4) - this is our destination!

Step 6: Mark destination as visited

• vis[4][4] = True
• Return from recursive calls

Final Result: After DFS completes, we check vis[4][4] which is True, so the function returns true - the ball can reach the destination.

The key insight: The ball doesn't stop at every cell. When rolling from (4,0) to the right, it passes through (4,1), (4,2), and (4,3) without stopping, finally stopping at (4,4) when it hits the right boundary.

Time and Space Complexity

Time Complexity: O(m * n * max(m, n))

The algorithm uses DFS to explore all reachable positions in the maze. In the worst case:

• We visit each cell at most once due to the visited array, giving us O(m * n) cells to explore
• For each cell we visit, we simulate rolling the ball in 4 directions
• Each roll simulation can take up to O(max(m, n)) steps in the worst case (rolling from one end of the maze to the other)
• Therefore, the total time complexity is O(m * n * max(m, n))

Space Complexity: O(m * n)

The space complexity consists of:

• The visited array vis which uses O(m * n) space to track visited cells
• The recursive call stack for DFS, which in the worst case can go up to O(m * n) depth if we visit all cells in a chain
• Therefore, the total space complexity is O(m * n)

Common Pitfalls

Pitfall 1: Confusing Visited Positions - Rolling vs. Stopping

The Problem: A common mistake is marking positions as visited while the ball is rolling through them, rather than only marking the positions where the ball stops. This leads to incorrect results because the ball might need to pass through the same position multiple times in different directions.

Incorrect Implementation:

def dfs(row, col):  if visited[row][col]:  return  visited[row][col] = True   for delta_row, delta_col in directions:  next_row, next_col = row, col  while (0 <= next_row + delta_row < rows and  0 <= next_col + delta_col < cols and  maze[next_row + delta_row][next_col + delta_col] == 0):  next_row += delta_row  next_col += delta_col  # WRONG: Marking intermediate positions as visited  visited[next_row][next_col] = True  dfs(next_row, next_col)

Why It's Wrong: Consider a simple maze where the ball needs to roll through position (2, 2) horizontally to reach (2, 4), then later roll through the same position (2, 2) vertically to reach the destination at (4, 2). If we mark (2, 2) as visited during the horizontal roll, the vertical path becomes blocked.

Correct Approach: Only mark positions as visited where the ball actually stops (when it hits a wall):

def dfs(row, col):  if visited[row][col]:  return  visited[row][col] = True # Only mark stopping position   for delta_row, delta_col in directions:  next_row, next_col = row, col  while (0 <= next_row + delta_row < rows and  0 <= next_col + delta_col < cols and  maze[next_row + delta_row][next_col + delta_col] == 0):  next_row += delta_row  next_col += delta_col  # No marking during rolling  dfs(next_row, next_col) # Only explore from stopping position

Pitfall 2: Incorrect Boundary Check in While Loop

The Problem: Using the wrong boundary check can cause index out of bounds errors or incorrect stopping positions.

Incorrect Implementation:

# WRONG: Checking current position instead of next position while (0 <= next_row < rows and 0 <= next_col < cols and  maze[next_row][next_col] == 0):  next_row += delta_row  next_col += delta_col

Why It's Wrong: This checks if the current position is valid and empty, but then moves to the next position without verifying it. The ball could roll into a wall or out of bounds.

Correct Approach: Always check the next position before moving:

while (0 <= next_row + delta_row < rows and  0 <= next_col + delta_col < cols and  maze[next_row + delta_row][next_col + delta_col] == 0):  next_row += delta_row  next_col += delta_col

Pitfall 3: Not Handling the Case Where Start Equals Destination

The Problem: If the starting position is the same as the destination, the solution should return True only if the ball can move away and come back to stop at that position. Simply returning True immediately would be incorrect.

Incorrect Implementation:

def hasPath(maze, start, destination):  if start == destination:  return True # WRONG: Assumes ball can stop without moving  # ... rest of the code

Why It's Wrong: The ball might be in the middle of an open area where it cannot stop. It must hit a wall to stop, so even if start equals destination, we need to verify that the ball can actually stop at that position.

Correct Approach: Let the DFS algorithm handle this case naturally. The ball will explore all possible paths, and if it can return to the starting position as a valid stopping point, the algorithm will mark it as visited:

def hasPath(maze, start, destination):  # No special case for start == destination  dfs(start[0], start[1])  return visited[destination[0]][destination[1]]

The algorithm correctly handles this because it only marks positions where the ball can actually stop (after hitting a wall).