Problem Description

In this problem, we are given a maze represented by a 2D matrix. Each cell of the matrix can either be an empty cell ('.') or a wall ('+'). An entrance to the maze is also given to us, specified by its row and column index. We are asked to find the shortest path from the entrance to the nearest exit of the maze. Here, an exit is defined as an empty cell located on the border of the maze, excluding the entrance itself. In each step, we can move to a cell that is adjacent (left, right, up, or down) to our current position, but we cannot move into walls or outside the maze boundaries.

Our goal is to navigate from the entrance to the nearest border cell (exit) by taking the fewest steps possible. If we can’t find any path to an exit, we must return -1. The number of steps is counted as the minimum number of moves required to reach any exit from the entrance.

Flowchart Walkthrough

Let's utilize the algorithm flowchart to analyze LeetCode 1926. Nearest Exit from Entrance in Maze and determine the most suitable approach using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The maze can be interpreted as a graph where each cell serves as a node, and edges exist between adjacent passages.

Is it a tree?

• No: The maze is not hierarchically structured as a tree and may include loops or multiple routes to a destination.

Is the problem related to directed acyclic graphs (DAGs)?

• No: The problem doesn't specify any directional constraints that prevent cycles, and loop backs are possible when wandering the maze.

Is the problem related to shortest paths?

• Yes: The task is to find the shortest path from an entrance to the nearest exit, classifying it as a shortest path problem.

Is the graph weighted?

• No: Moving from one maze cell to a neighboring cell is typically of equal cost, indicating an unweighted graph.

Conclusion: By following the flowchart steps, the most appropriate strategy for this unweighted, shortest path problem in a non-DAG structure is Breadth-First Search (BFS). This approach is well-suited to exploring levels of the maze systematically and finding the shortest route to the nearest exit.

Intuition

To solve this problem, we can use the Breadth-First Search (BFS) algorithm. BFS is an ideal choice for this kind of problem because it explores all possible paths level by level or, in this case, step by step from the entrance. As a result, the first time it reaches an exit, it is guaranteed to be the nearest one since BFS doesn't explore deeper paths until all paths of the current depth are explored.

We initialize BFS from the entrance by marking it as visited (to avoid revisiting) and then iteratively exploring all four adjacent cells. If an adjacent cell is empty and within the maze bounds, we check if it's an exit. If it's an exit, we immediately return the current step count since it's the minimum. If it's not, we continue the BFS by adding the cell to the queue. Importantly, as we enqueue a cell, we mark it with a wall to avoid revisiting cells that are already considered, effectively reducing unnecessary calculations.

If the BFS completes without finding an exit, we conclude that no path exists, and we return -1, indicating failure to reach an exit.

Learn more about Breadth-First Search patterns.

Solution Approach

The solution provided uses the Breadth-First Search (BFS) algorithm to traverse through the maze. Let's dissect the given Python code to better understand how it translates the BFS strategy into a working solution.

class Solution:
    def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
        m, n = len(maze), len(maze[0])  # Dimensions of the maze
        i, j = entrance  # Starting position (entrance of the maze)
        q = deque([(i, j)])  # Initialize the queue with the entrance
        maze[i][j] = '+'  # Mark the entrance as visited by replacing it with '+'
        ans = 0  # Steps counter
        while q:
            ans += 1  # Increment the step counter at the beginning of each level
            for _ in range(len(q)):  # Loop over every node in the current level
                i, j = q.popleft()  # Dequeue the front element from the queue
                for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:  # The 4 possible directions
                    x, y = i + a, j + b  # Calculate the next cell's coordinates
                    if 0 <= x < m and 0 <= y < n and maze[x][y] == '.':  # Check if it's within bounds and not visited
                        if x == 0 or x == m - 1 or y == 0 or y == n - 1:  # Check if it's an exit
                            return ans  # Return the current step count as the answer
                        q.append((x, y))  # Enqueue the cell for future exploration
                        maze[x][y] = '+'  # Mark the cell as visited
        return -1  # If the loop ends without finding an exit, return -1

Key Elements of the Solution:

• Queue (deque): A queue is used for the BFS traversal, which follows the First-In-First-Out (FIFO) principle. This ensures that cells are explored in the order they are reached.

• Visited Marking: When a cell is visited, it is marked by changing its value to '+'. This prevents the algorithm from re-visiting the same cell, which would otherwise lead to infinite loops and incorrect step count.

• Direction Array: A simple 2D array [[0, -1], [0, 1], [-1, 0], [1, 0]] is used to represent the four possible moves from any cell (left, right, up, down).

• Checking Exit Condition: An exit is an empty cell ('.') at the border of the maze. Whenever moving to a new cell, the algorithm checks if it is an exit by comparing its coordinates with the boundary of the maze.

• Steps Counter: The variable ans is used to count the number of steps taken to reach an exit. It gets incremented at the start of processing each level, ensuring the correct step count when an exit is found.

Efficiency of the Solution:

The solution is efficient for finding the shortest path in a maze scenario. BFS ensures that the first time we find an exit, it must be the closest one because we explore all possible paths of equal length before moving on to longer paths. The time and space complexity of this solution are both O(m*n), where m is the number of rows and n is the number of columns in the maze.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach. Consider the following 5x5 maze as an example, where E is the entrance, . represents an empty cell, and + represents a wall:

+ + + + +
+ . . . +
+ . E . +
+ . . . +
+ + + . +

Following the BFS strategy to navigate the maze from the entrance to the closest exit:

1. Initialize the queue with the entrance coordinates, which is (2, 2) here, and set the starting cell as visited by marking it with '+':

+ + + + +
+ . . . +
+ . + . +
+ . . . +
+ + + . +

2. Process the first node in the queue, we explore its adjacent cells: (2, 1), (2, 3), (1, 2), and (3, 2). These cells are not yet visited; we add them to the queue and mark them as visited:

+ + + + +
+ . + . +
+ + + . +
+ . + . +
+ + + . +

3. We increment our steps counter ans = 1 as we have started moving from the entrance. Now we process the cells in the queue. For each cell, we check its adjacent cells. For example, start with cell (2, 1) and check (1, 1), (3, 1), (2, 0), and (2, 2). Here, (2, 0) is not within the bounds, (2, 2) is already visited, and (1, 1) and (3, 1) are available for further exploration.

4. As we continue to explore the current level, we find that the cell (1, 2) is just one step away from an exit as it is located on the border of the maze. We've reached an exit:

+ + + + +
+ . + . +
+ + + . +
+ . + . +
+ + + . +

5. Since we've encountered an exit, we immediately return the current step count, which is ans = 1.

By the BFS approach, we've managed to find the nearest exit to the entrance in the fewest steps as possible without unnecessary calculations. If no exit had been reached, we would have continued to process the BFS queue until it was empty and then returned -1 to indicate no available exit.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(M*N), where M is the number of rows and N is the number of columns in the maze. This complexity arises because, in the worst-case scenario, the algorithm needs to traverse all the cells in the maze before finding an exit or determining that no exit is reachable. Traversal is done via breadth-first search (BFS), and each cell is visited at most once, as visited cells are marked with '+' to prevent revisiting.

The space complexity is also O(M*N) due to the same reasoning. The space is used to store the queue q, which, in the worst case, could contain all the cells as we explore the maze. Furthermore, we modify the maze to mark visited positions, which also takes O(M*N) space; however, this does not add to the complexity as it is the same maze that is passed as input and not additional space being used.

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