Paths in Maze That Lead to Same Room

Let's explore how to determine the number of paths in a maze that lead to the same room.

Naive Approach: Brute Force

1. Description:

   • The most straightforward approach is to explore all possible paths in the maze.
   • For each path, check the destination room.
   • Count the paths that end up in the same room.

2. Algorithm:

   • Use Depth-First Search (DFS) or Breadth-First Search (BFS) to traverse all possible paths.
   • Maintain a count for each room.
   • Increment the count for the room when a path ends in that room.
   • Finally, calculate the number of pairs of paths that lead to the same room. If count[i] is the number of paths that lead to room i, then the number of pairs is count[i] * (count[i] - 1) / 2.

3. Big O Analysis:

   • Time Complexity: O(2^N) in the worst-case scenario, where N is the number of cells in the maze, since each cell can have multiple paths to explore.
   • Space Complexity: O(N) for the recursion stack or queue in DFS/BFS.

4. Code (Python):

def count_paths_brute_force(maze, start, end):
    rows, cols = len(maze), len(maze[0])
    count = 0

    def dfs(row, col, path):
        nonlocal count
        if row < 0 or row >= rows or col < 0 or col >= cols or maze[row][col] == '#' or (row, col) in path:
            return

        if row == end[0] and col == end[1]:
            count += 1
            return

        path.add((row, col))
        dfs(row + 1, col, path.copy())
        dfs(row - 1, col, path.copy())
        dfs(row, col + 1, path.copy())
        dfs(row, col - 1, path.copy())
        path.remove((row, col))

    dfs(start[0], start[1], set())
    return count


maze = [
    ['.', '.', '.'],
    ['.', '#', '.'],
    ['.', '.', '.']
]
start = (0, 0)
end = (2, 2)

num_paths = count_paths_brute_force(maze, start, end)
print(f"Number of paths from {start} to {end}: {num_paths}")

Optimal Approach: Dynamic Programming (If Applicable)

1. Description:

   • If the maze structure allows for it, Dynamic Programming (DP) can be employed to optimize the solution by avoiding redundant calculations.
   • DP is particularly effective when the maze has overlapping subproblems.

2. Algorithm:

   • Create a DP table dp[i][j][k] where dp[i][j][k] represents the number of paths to cell (i, j) with length k.
   • The base case is dp[start_row][start_col][0] = 1.
   • Iteratively compute the number of paths to each cell based on its neighboring cells.
   • The result would be to count pairs of paths which end in the same cell (i, j) and potentially have the same length k.

3. Big O Analysis:

   • Time Complexity: O(R * C * L), where R and C are the dimensions of the maze, and L is the maximum path length.
   • Space Complexity: O(R * C * L) to store the DP table.

4. Edge Cases:

   • Empty Maze: If the maze is empty or invalid, return 0.
   • No Path: If there is no path from the start to the end, return 0.
   • Obstacles: Handle obstacles in the maze appropriately.
   • Start and End Same: If the start and end points are the same, there is at least one path of length 0.
   • Large Maze: For a very large maze, consider memory constraints and potential optimizations like using rolling arrays.

5. Code (Python - Dynamic Programming):

def count_paths_dp(maze, start, end, max_length):
    rows, cols = len(maze), len(maze[0])
    dp = {}  # dp[(row, col, length)] = number of paths

    def solve(row, col, length):
        if row < 0 or row >= rows or col < 0 or col >= cols or maze[row][col] == '#' or length > max_length:
            return 0

        if (row, col, length) in dp:
            return dp[(row, col, length)]

        if row == end[0] and col == end[1]:
            dp[(row, col, length)] = 1
            return 1

        dp[(row, col, length)] = (
            solve(row + 1, col, length + 1) +
            solve(row - 1, col, length + 1) +
            solve(row, col + 1, length + 1) +
            solve(row, col - 1, length + 1)
        )

        return dp[(row, col, length)]

    total_paths = solve(start[0], start[1], 0)
    return total_paths


maze = [
    ['.', '.', '.'],
    ['.', '#', '.'],
    ['.', '.', '.']
]
start = (0, 0)
end = (2, 2)
max_length = 10  # Maximum path length to consider

num_paths = count_paths_dp(maze, start, end, max_length)
print(f"Number of paths from {start} to {end}: {num_paths}")