The Maze III

Updated on 04 July, 2025
The Maze III header image

Problem Statement

A ball is placed in a maze consisting of empty spaces (denoted by 0) and walls (indicated by 1). The ball moves through the maze in four cardinal directions - up, down, left, and right, but it can only stop when it hits a wall. At each stopping point, it can choose a new direction to continue. Within such a setup, a hole exists in the maze, and the objective is for the ball to end its journey in this hole.

Given the dimensions of the maze m x n, the starting position of the ball, and the location of the hole, the task is to compute the shortest possible route or sequence of moves the ball can take to reach the hole. This route has to be provided in the form of a string, instructions, composed of the letters u (up), d (down), l (left), and r (right). The string should indicate the shortest series of moves to get the ball into the hole, or return "impossible" if no such path exists. If there are paths tieing in distance, return the lexicographically first path. The distance is calculated based on the number of empty spaces covered by the ball from its start position to the hole, excluding the ball's start space and including the destination space.

The maze itself is bordered by walls, reinforcing the complexity of the movement.

Examples

Example 1

Input:

maze = [[0,0,0,0,0],[1,1,0,0,1],[0,0,0,0,0],[0,1,0,0,1],[0,1,0,0,0]], ball = [4,3], hole = [0,1]

Output:

"lul"

Explanation:

There are two shortest ways for the ball to drop into the hole.
The first way is left -> up -> left, represented by "lul".
The second way is up -> left, represented by 'ul'.
Both ways have shortest distance 6, but the first way is lexicographically smaller because 'l' < 'u'. So the output is "lul".

Example 2

Input:

maze = [[0,0,0,0,0],[1,1,0,0,1],[0,0,0,0,0],[0,1,0,0,1],[0,1,0,0,0]], ball = [4,3], hole = [3,0]

Output:

"impossible"

Explanation:

The ball cannot reach the hole.

Example 3

Input:

maze = [[0,0,0,0,0,0,0],[0,0,1,0,0,1,0],[0,0,0,0,1,0,0],[0,0,0,0,0,0,1]], ball = [0,4], hole = [3,5]

Output:

"dldr"

Constraints

  • m == maze.length
  • n == maze[i].length
  • 1 <= m, n <= 100
  • maze[i][j] is 0 or 1.
  • ball.length == 2
  • hole.length == 2
  • 0 <= ballrow, holerow <= m
  • 0 <= ballcol, holecol <= n
  • Both the ball and the hole exist in an empty space, and they will not be in the same position initially.
  • The maze contains at least 2 empty spaces.

Approach and Intuition

The problem of guiding the ball to the hole can be visualized as a pathfinding problem in a weighted grid, where movement can continue in one direction until obstructed:

  1. Utilize Breadth-First Search (BFS) or Dijkstra's Algorithm:

    • Treat each valid position in the maze as a node in a graph.
    • Movement directions (u, d, l, r) represent the edges, with weights being the distance traveled in one continuous movement until hitting a wall or falling into the hole.
  2. Determine movement constraints from the maze setup:

    • The ball stops only when hitting walls, indicating the end of one path segment.
    • Update the minimum distance to a new point each time it is reached.
  3. Implement a priority queue for effective BFS:

    • Priority is given to paths with the shortest distance.
    • To handle lexicographical order in paths of the same length, use a lexical comparison as a secondary criterion in the queue.
  4. Continuously update and check paths:

    • From a given point, explore all four possible directions.
    • For each direction, move as far as possible until hitting a wall or the boundary of the maze, or dropping into the hole.
    • Update the path if this new point has a shorter path than previously recorded.
  5. Cease exploration when:

    • The hole is reached by the ball. Here, check if the reached condition satisfies the shortest path criteria.
    • If exploration completes without reaching the hole, return "impossible".

From the example scenarios provided:

  • Example 1 demonstrates a case where multiple paths exist, and the shortest lexicographical string is chosen.
  • Example 2 showcases an impossible situation with no viable path to the destination.
  • Example 3 illustrates a direct pathing using a combination of movements to reach the target effectively.

Through careful consideration of direction, path length, and each cell visited, an optimal solution efficiently leads the ball to its goal.

Solutions

  • Java
java
class Position {
    int x;
    int y;
    int steps;
    String route;
        
    public Position(int x, int y, int steps, String route) {
        this.x = x;
        this.y = y;
        this.steps = steps;
        this.route = route;
    }
}
    
class Pathfinder {
    int[][] moves = new int[][]{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    String[] directions = new String[]{"r", "d", "l", "u"};
    int rows;
    int cols;
        
    public String shortestPath(int[][] labyrinth, int[] start, int[] target) {
        rows = labyrinth.length;
        cols = labyrinth[0].length;
            
        PriorityQueue<Position> queue = new PriorityQueue<>((p1, p2) -> {
            int compareDist = Integer.compare(p1.steps, p2.steps);
            if (compareDist == 0) {
                return p1.route.compareTo(p2.route);
            }
            return compareDist;
        });
            
        boolean[][] visited = new boolean[rows][cols];
        queue.add(new Position(start[0], start[1], 0, ""));
            
        while (!queue.isEmpty()) {
            Position current = queue.poll();
            int x = current.x;
            int y = current.y;
                
            if (visited[x][y]) {
                continue;
            }
                
            if (x == target[0] && y == target[1]) {
                return current.route;
            }
                
            visited[x][y] = true;
                
            for (Position neighbor: getAdjacent(x, y, labyrinth, target)) {
                int nextX = neighbor.x;
                int nextY = neighbor.y;
                int nextSteps = neighbor.steps;
                String move = neighbor.route;
                queue.add(new Position(nextX, nextY, current.steps + nextSteps, current.route + move));
            }
        }
            
        return "impossible";
    }
        
    private boolean isAccessible(int x, int y, int[][] labyrinth) {
        return x >= 0 && x < rows && y >= 0 && y < cols && labyrinth[x][y] == 0;
    }
        
    private List<Position> getAdjacent(int x, int y, int[][] labyrinth, int[] target) {
        List<Position> adjacent = new ArrayList<>();
            
        for (int i = 0; i < 4; i++) {
            int dx = moves[i][0];
            int dy = moves[i][1];
            String moveDirection = directions[i];
                
            int currX = x;
            int currY = y;
            int distance = 0;
                
            while (isAccessible(currX + dx, currY + dy, labyrinth)) {
                currX += dx;
                currY += dy;
                distance++;
                if (currX == target[0] && currY == target[1]) {
                    break;
                }
            }
                
            adjacent.add(new Position(currX, currY, distance, moveDirection));
        }
            
        return adjacent;
    }
}

In the provided Java code for solving "The Maze III," the main focus is finding the shortest path through a maze represented by a grid using the Position and Pathfinder classes. The maze positions are either passable or blocked, depicted by 0s and 1s respectively, in the labyrinth 2D array.

The Position class keeps track of a coordinate (x, y), the number of steps taken to reach this point, and the path taken as a string of directions. Each instance of Position represents a state in the maze exploration process.

The Pathfinder class provides the method shortestPath to determine the shortest route from the start point to the target in the maze. The method uses a priority queue to manage states based on the fewest steps first, and if steps are equal, lexicographical order of the path.

The algorithm involves the following steps:

  1. Initialize variables for the number of rows and columns in the labryinth.
  2. Create a priority queue to manage exploring positions based on the least steps and lexicographical order of paths if steps are the same.
  3. Mark the start position as visited initially.
  4. Use a while loop to continually extract the position with the highest priority (least number of steps or best lexicographical order) and attempt to explore further from that point.
    • If the current position is already visited, it is skipped.
    • If the position is the target, the path to this position is returned.
    • Otherwise, adjacent positions (possible moves from the current position) are calculated. For each adjacent position not blocked and not revisited, it is added to the priority queue for further exploration.
  5. If the queue is exhausted without finding the target, the method returns "impossible" indicating no path exists.

Adjacency and movement are determined by the getAdjacent method, which checks all four possible moves (up, down, left, right) from a current position, considering only in-bounds and unblocked directions.

The overall solution thus utilizes systematic exploration with priority and lexicographical ordering for optimal and deterministic pathfinding in a maze scenario.

  • Python
python
class Solution:
    def findMinPath(self, labyrinth: List[List[int]], start: List[int], end: List[int]) -> str:
        def is_free(r, c):
            return 0 <= r < height and 0 <= c < width and labyrinth[r][c] == 0
            
        def explore(r, c):
            directions = [(0, 1, 'r'), (0, -1, 'l'), (1, 0, 'd'), (-1, 0, 'u')]
            possible_moves = []
                
            for dr, dc, dir in directions:
                new_r, new_c, travel = r, c, 0
                    
                while is_free(new_r + dr, new_c + dc):
                    new_r += dr
                    new_c += dc
                    travel += 1
                    if [new_r, new_c] == end:
                        break
                         
                possible_moves.append((new_r, new_c, travel, dir))
                    
            return possible_moves
    
        height = len(labyrinth)
        width = len(labyrinth[0])
        priority_queue = [(0, "", start[0], start[1])]
        visited = set()
            
        while priority_queue:
            dist, seq, r, c = heappop(priority_queue)
                
            if (r, c) in visited:
                continue
                
            if [r, c] == end:
                return seq
                
            visited.add((r, c))
                
            for new_r, new_c, travel_dist, dir in explore(r, c):
                heappush(priority_queue, (dist + travel_dist, seq + dir, new_r, new_c))
    
        return "impossible"

The task in this solution involves navigating through a maze to find the shortest path from a start point to an end point using Python 3 code. Here’s an overview of the solution steps implemented in the provided code:

  • Define the findMinPath method in the Solution class, which accepts the maze structure (labyrinth), a start point, and an end point. This method is aimed at finding the lexicographically smallest direction string for the shortest path.

  • Define helper function is_free(r, c) to check if a cell at row r and column c is navigable (0 means free and 1 means blocked).

  • Establish a utility function explore(r, c), which evaluates all possible moves from the current position and returns these moves along with their respective directions and distances.

  • Initialize labyrinth dimensions with height and width.

  • Use a priority queue (priority_queue) to manage exploration fronts, initialized with the starting position, path cost set to zero, and an empty direction sequence.

  • Add a visited set to keep track of already explored positions to avoid revisits and infinite loops.

  • Employ a priority queue mechanism, where the next position to explore is decided based on the accumulated distance (priority) and lexicographical order of path directions.

  • For each position, check its possibilities by calling explore(r, c). For each potential move explored:

    • Compute the total distance for the move.
    • Record the path sequentially in the form of direction letters.
    • Push the updated state back to the priority queue for further exploration.
  • If the end position is reached, return the accumulated direction sequence as the result.

  • If no path exists to reach the destination, return "impossible" as the final output.

This structured approach ensures that the shortest path, considering both path length and lexicographical order of movements, is found efficiently using a combination of BFS (Breadth-First Search) with a priority queue and lexicographical ordering facilitated through string manipulations.

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