Nearest Exit from Entrance in Maze

Medium

Nearest Exit from Entrance in Maze

Medium

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the maze (number of rows and columns), and what are the maximum possible values for these dimensions?
What values represent an open cell versus a wall in the maze array? Is it guaranteed to be ' ' (empty space) and '+' (plus sign), or could other characters be used?
Is it guaranteed that the entrance point will always be a valid open cell within the maze?
If there is no exit reachable from the entrance, what should the function return? (e.g., -1, null, or throw an exception?)
Can the entrance be located directly at an exit? If so, what should be returned?

Brute Force Solution

Approach

Imagine you're standing at the entrance of a maze, and you want to find the nearest exit. The brute force approach means you try every single possible path from the entrance until you find an exit, and then pick the shortest path among all the paths that lead to an exit.

Here's how the algorithm would work step-by-step:

Start at the entrance.
Try going in every possible direction (up, down, left, right).
For each direction you move, check if you've reached an exit.
If you have reached an exit, remember the number of steps it took to get there.
If you haven't reached an exit, try going in every possible direction again from your new location.
Keep doing this, exploring every possible path, until you've explored every single path from the entrance or you have reached all exits.
After exploring all paths, compare the number of steps for each exit you found and choose the path with the fewest steps.

Code Implementation

def nearest_exit_brute_force(maze, entrance):
    rows = len(maze)
    cols = len(maze[0])
    entrance_row = entrance[0]
    entrance_col = entrance[1]
    shortest_distance = float('inf')

    def is_valid(row, col):
        return 0 <= row < rows and 0 <= col < cols and maze[row][col] == '.'

    def is_exit(row, col):
        return (row == 0 or row == rows - 1 or col == 0 or col == cols - 1) and (row, col) != (entrance_row, entrance_col)

    def explore_paths(row, col, distance, visited):
        nonlocal shortest_distance

        if is_exit(row, col):
            shortest_distance = min(shortest_distance, distance)
            return

        # Mark current cell as visited to avoid cycles.
        visited.add((row, col))

        # Define possible movement directions: up, down, left, right.
        directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

        for row_direction_change, col_direction_change in directions:
            new_row = row + row_direction_change
            new_col = col + col_direction_change

            # Only explore if new cell is valid and unvisited.
            if is_valid(new_row, new_col) and (new_row, new_col) not in visited:

                explore_paths(new_row, new_col, distance + 1, visited.copy())

    explore_paths(entrance_row, entrance_col, 0, set())

    if shortest_distance == float('inf'):
        return -1
    else:
        return shortest_distance

Big(O) Analysis

Time Complexity

O(4^(m*n)) – The brute force approach explores every possible path in the maze. In the worst-case scenario, the algorithm might visit each cell multiple times by going back and forth. From each cell, there are up to 4 possible directions to move. If the maze has dimensions m x n, then the maximum possible path length is proportional to m*n. Since the algorithm explores all possible paths of varying lengths up to the maximum path length with up to 4 choices at each step, the total number of paths explored can grow exponentially. Therefore, the time complexity is O(4^(m*n)).

Space Complexity

O(N) – The brute force approach explores all possible paths in the maze. In the worst-case scenario, the algorithm might need to keep track of all cells in the maze to remember the path it has explored or will explore. This tracking is typically done using a queue or a stack (implicit data structures during the exploration process) for breadth-first search (BFS) or depth-first search (DFS) style traversals. Therefore, the space used for storing this path information can grow linearly with the number of cells in the maze, which we denote as N. Hence, the space complexity is O(N).

Optimal Solution

Approach

The key to solving this maze problem efficiently is to explore the maze in a breadth-first manner, like ripples spreading across water. This ensures that we find the shortest path to any exit before considering longer paths. This helps avoid unnecessary exploration of longer paths.

Here's how the algorithm would work step-by-step:

Start at the entrance of the maze.
Imagine expanding outwards from the entrance, step by step, visiting all neighboring empty cells at each step.
Keep track of how many steps it took to reach each cell from the entrance.
The first time you encounter an exit cell, you know you've found the shortest path to an exit.
The number of steps it took to reach that exit cell is the length of the shortest path.

Code Implementation

def nearest_exit(maze, entrance):
    rows = len(maze)
    columns = len(maze[0])
    queue = [(entrance[0], entrance[1], 0)]
    visited = {tuple(entrance)}

    # Possible directions to move (up, down, left, right).
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    while queue:
        row_index, column_index, steps = queue.pop(0)

        # Check if current cell is an exit
        if (row_index == 0 or row_index == rows - 1 or
            column_index == 0 or column_index == columns - 1) and\
           (row_index, column_index) != (entrance[0], entrance[1]):

            return steps

        for row_direction, column_direction in directions:
            new_row_index = row_index + row_direction
            new_column_index = column_index + column_direction

            # Only explore valid and unvisited cells.
            if (0 <= new_row_index < rows and
                0 <= new_column_index < columns and
                maze[new_row_index][new_column_index] == '.' and
                (new_row_index, new_column_index) not in visited):

                # Mark as visited to avoid cycles.
                visited.add((new_row_index, new_column_index))

                # Add valid cells to the queue for exploration
                queue.append((new_row_index, new_column_index, steps + 1))

    # No exit found.
    return -1

Big(O) Analysis

Time Complexity

O(m*n) – The algorithm uses breadth-first search (BFS) to explore the maze. In the worst-case scenario, we might need to visit every empty cell in the maze. If the maze has dimensions m x n, where m is the number of rows and n is the number of columns, the total number of cells is m*n. The BFS algorithm visits each cell at most once. Therefore, the time complexity is proportional to the number of cells in the maze, resulting in O(m*n).

Space Complexity

O(M*N) – The provided solution utilizes Breadth-First Search (BFS). The primary auxiliary space is used by the queue, which stores the cells to be visited. In the worst-case scenario, where the maze is mostly empty, the queue could potentially hold all the cells in the maze, leading to a space complexity proportional to the maze's dimensions. Therefore, the space complexity is O(M*N), where M is the number of rows and N is the number of columns in the maze, representing the maximum number of cells stored in the queue.

Edge Cases

Case	How to Handle
Null or empty maze input	Return -1 immediately as there's no maze to traverse.
Maze with only one cell (1x1)	If the entrance is not an exit, return -1; otherwise, return 0.
Entrance is on the border of the maze (already an exit)	Return 0 immediately as the entrance itself is the nearest exit.
Maze has no exits (entrance is surrounded by walls)	The BFS should terminate without finding any exits, and the algorithm should return -1.
Entrance is blocked (surrounded by walls)	The BFS will not be able to start and will return -1 since no path is possible.
Large maze dimensions (potential memory issues)	BFS ensures that memory usage scales linearly with the number of cells visited; avoid recursion to prevent stack overflow.
Maze with disconnected components (entrance is in a component with no exits)	The BFS will explore the connected component and terminate without finding any exits, returning -1.
Integer overflow potential with large maze dimensions and long path lengths	Using `int` for the distance should be sufficient, but consider using `long` if extremely large mazes are possible to avoid potential overflow.