Paths in Maze That Lead to Same Room

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GraphsDynamic Programming

You are given a 0-indexed n x n integer matrix maze representing the maze. The maze has walls, empty cells, and start/end cells. You can go from a cell to another cell in one step if they are adjacent in one of the four directions and the destination cell is not a wall.

More formally, in each step, you can move from cell (r1, c1) to cell (r2, c2) if:

r1 == r2 and abs(c1 - c2) == 1
c1 == c2 and abs(r1 - r2) == 1
The cell (r2, c2) is not a wall.

You are also given a 0-indexed integer array entrance where entrance = [entrance_row, entrance_col] represents the location of the maze entrance. You are also given a 0-indexed integer array exit where exit = [exit_row, exit_col] represents the location of the maze exit.

You need to add one or more new rooms so that there is a path from the entrance to the exit such that this path visits all the rooms in the maze exactly once. A room is a non-wall cell in the maze.

Return the minimum number of rooms to add such that there is a path from the entrance to the exit such that this path visits all the rooms in the maze exactly once. It can be proven that there is always an answer.

Notes:

The entrance and exit could be the same cell.
You can create rooms in any place in the maze as long as they are not in the same location as a wall.

Example 1:

Input: maze = [[0,0,0],[1,0,0],[1,1,0]], entrance = [0,0], exit = [1,1]
Output: 0
Explanation: We have already a path that visits all the rooms.

Example 2:

Input: maze = [[0,0,0],[1,0,0],[1,1,0]], entrance = [0,0], exit = [2,2]
Output: 1
Explanation: You need to add one room so that there is a path that visits all the rooms.

Example 3:

Input: maze = [[0,0,0],[1,0,0],[1,1,0]], entrance = [1,1], exit = [1,1]
Output: 2
Explanation: You need to add two rooms so that there is a path that visits all the rooms.

Constraints:

n == maze.length == maze[i].length
2 <= n <= 100
maze[i][j] is 0 or 1.
entrance.length == exit.length == 2
0 <= entrance_row, entrance_col, exit_row, exit_col < n

Try coding it in LeetCode

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 (rows x columns)? Are they constrained to a certain range?
What values represent walls, empty paths, and the 'same room'? Are we using integers, booleans, or another data type?
Can the maze have cycles, and if so, should I handle paths that revisit the same cell multiple times?
If there are multiple paths leading to the same room, do I need to return all of them, the shortest one, or just the count?
If there is no path that leads to the same room, what should the function return (e.g., null, empty list, or a specific error code)?

Brute Force Solution

Approach

The problem asks us to find different ways to navigate a maze and end up in the same room. The brute force way means we'll try every single possible path, no matter how long or winding, and see if it works.

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

Start at the maze's starting point.
Explore every possible direction you can move from the starting point.
For each of those moves, explore every possible direction you can move from that new room.
Keep going, making all possible moves at each room, until you either reach the room we're looking for or get stuck.
If you reach the target room, mark that path as a valid one.
If you get stuck (hit a dead end), discard that path and try a different route.
Repeat steps 2-6 until you have tried every possible path through the maze.
Count up how many of those paths led to the room we were looking for. That's your answer.

Code Implementation

def find_paths_brute_force(maze, start_node, end_node):
    number_of_paths = 0

    def explore_paths(current_node, current_path):
        nonlocal number_of_paths

        # If we've reached the end, increment the count.
        if current_node == end_node:

            number_of_paths += 1
            return

        # If we hit a dead end, terminate the current path
        if not maze[current_node]:

            return

        # Explore all adjacent nodes.
        for next_node in maze[current_node]:

            # Build the new path.
            new_path = current_path + [next_node]
            explore_paths(next_node, new_path)

    explore_paths(start_node, [start_node])
    return number_of_paths

Big(O) Analysis

Time Complexity

O(4^n) – The given solution employs a brute-force approach to explore all possible paths in the maze. In the worst-case scenario, where the maze has few or no obstacles, each room could have up to four possible directions to move (up, down, left, right). The algorithm recursively explores each of these directions. If 'n' represents the number of rooms in the maze, the depth of the recursion can go up to 'n', and at each level of the recursion, there are up to 4 choices. Therefore, the total number of possible paths explored grows exponentially with the size of the maze, resulting in a time complexity of approximately O(4^n).

Space Complexity

O(N) – The brute-force approach described uses recursion to explore all possible paths. The depth of the recursion can be at most N, where N is the number of cells in the maze, in the worst-case scenario where a path visits every cell. Each recursive call requires space on the call stack to store function arguments and local variables. Therefore, the auxiliary space complexity is proportional to the maximum recursion depth, resulting in O(N) space.

Optimal Solution

Approach

The core idea is to explore the maze in a systematic way, keeping track of where we've been. We want to find different paths that eventually lead us to the same spot in the maze, representing revisiting a room from different directions.

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

Imagine you're exploring the maze. You start at the entrance and mark each room you visit.
As you move, keep track of the exact route you took to get to each room. Think of it like writing down your steps.
If you arrive at a room you've already visited, check if the path you took to get there this time is different from the previous path.
If the new path is different, you've found two distinct ways to reach the same room! Count this as a successful discovery.
Keep exploring the maze, remembering the routes and looking for those repeated rooms with different paths. This way, you can find all the unique pairs of paths that lead to the same room.

Code Implementation

def find_paths_to_same_room(maze):
    rows = len(maze)
    cols = len(maze[0]) if rows > 0 else 0
    paths_count = 0
    visited = set()
    room_paths = {}

    def explore_maze(row_index, col_index, current_path):
        nonlocal paths_count

        if row_index < 0 or row_index >= rows or col_index < 0 or col_index >= cols or maze[row_index][col_index] == '#' or (row_index, col_index, tuple(current_path)) in visited:
            return

        visited.add((row_index, col_index, tuple(current_path)))

        if (row_index, col_index) in room_paths:
            # Check if the current path is different from existing paths
            existing_paths = room_paths[(row_index, col_index)]
            if current_path not in existing_paths:
                paths_count += 1
                room_paths[(row_index, col_index)].append(current_path)
        else:
            # First time visiting this room, record the path.
            room_paths[(row_index, col_index)] = [current_path]

        # Explore all possible directions from the current room.
        explore_maze(row_index + 1, col_index, current_path + ['D'])

        explore_maze(row_index - 1, col_index, current_path + ['U'])

        explore_maze(row_index, col_index + 1, current_path + ['R'])

        explore_maze(row_index, col_index - 1, current_path + ['L'])

    # Start exploring from the entrance (0, 0) with an empty path.
    explore_maze(0, 0, [])

    return paths_count

Big(O) Analysis

Time Complexity

O(m * n) – Let m be the number of cells in the maze and n be the maximum path length from the starting point. Exploring the maze involves visiting each cell (m) and, for each cell, we might need to store or compare paths of maximum length n. In the worst case, we explore each possible path up to length n for each cell. Comparing paths could involve up to n operations. Therefore, the overall time complexity is O(m * n), where m represents the number of cells and n represents the maximum path length.

Space Complexity

O(N^2) – The algorithm keeps track of visited rooms and the path taken to reach each room. This requires storing, for each visited room, the sequence of steps (path) taken to get there. In the worst-case scenario, we might visit every room in the maze (let's say there are N rooms, representing the total size of the maze). For each of these N rooms, we may need to store a path of length up to N, leading to a space complexity of potentially O(N*N) to track these paths. Additionally, a set or hashmap might be used to store these paths, contributing further to space. Therefore, the space complexity is approximately O(N^2).

Edge Cases

Case	How to Handle
Null or empty maze (grid)	Return an empty list as there are no paths possible in an empty maze.
Maze with dimensions 1x1 (single cell)	Return an empty list as a path of length > 1 is impossible so reaching same room twice is impossible.
Maze with no path to any room (all walls)	Return an empty list because no room can be reached twice.
Maze with one room accessible from starting point	If the only accessible room is the start, return an empty list since we can't return to it via a different path.
Large maze where the search could result in stack overflow if recursion is not handled properly	Use iterative DFS or BFS to avoid stack overflow issues in very large mazes.
Maze with cycles where path length exceeds the maximum allowed path length	Implement cycle detection or path length limitation to prevent infinite loops.
Maze where all paths lead to the same room	The algorithm should correctly identify that all possible paths end in the same room and return this room.
Maze containing integer overflow or very large coordinate values	Use appropriate data types (e.g., long) or modulo operations to avoid overflow issues when calculating coordinates or path lengths.

Null or empty maze (grid)

How to Handle:

Return an empty list as there are no paths possible in an empty maze.

Maze with dimensions 1x1 (single cell)

How to Handle:

Return an empty list as a path of length > 1 is impossible so reaching same room twice is impossible.

Maze with no path to any room (all walls)

How to Handle:

Return an empty list because no room can be reached twice.

Maze with one room accessible from starting point

How to Handle:

If the only accessible room is the start, return an empty list since we can't return to it via a different path.

Large maze where the search could result in stack overflow if recursion is not handled properly

How to Handle:

Use iterative DFS or BFS to avoid stack overflow issues in very large mazes.

Maze with cycles where path length exceeds the maximum allowed path length

How to Handle:

Implement cycle detection or path length limitation to prevent infinite loops.

Maze where all paths lead to the same room

How to Handle:

The algorithm should correctly identify that all possible paths end in the same room and return this room.

Maze containing integer overflow or very large coordinate values

How to Handle:

Use appropriate data types (e.g., long) or modulo operations to avoid overflow issues when calculating coordinates or path lengths.