The Maze II

There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up,down,left or right, but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction.

Given the ball's start position, the destination and the maze, find the shortest distance for the ball to stop at the destination. The distance is defined by the number of empty spacestraveled by the ball from the start position (excluded) to the destination (included). If the ball cannot stop at the destination, return -1.

The maze is represented by a binary 2D array. 1 means the wall and 0 means the empty space. You may assume that the borders of the maze are all walls. The start and destination coordinates are represented by row and column indexes.

Example 1:

Input 1: a maze represented by a 2D array

0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0

Input 2: start coordinate (rowStart, colStart) = (0, 4)
Input 3: destination coordinate (rowDest, colDest) = (4, 4)

Output: 12

Explanation: One shortest way is : left -> down -> left -> down -> right -> down -> right.
             The total distance is 1 + 1 + 3 + 1 + 2 + 2 + 2 = 12.

Example 2:

Input 1: a maze represented by a 2D array

0 0 1 0 0
0 0 0 0 0
0 0 0 1 0
1 1 0 1 1
0 0 0 0 0

Input 2: start coordinate (rowStart, colStart) = (0, 4)
Input 3: destination coordinate (rowDest, colDest) = (3, 2)

Output: -1

Explanation: There is no way for the ball to stop at the destination.

Note:

1. There is only one ball and one destination in the maze.

2. Both the ball and the destination exist on an empty space, and they will not be at the same position initially.

3. The given maze does not contain border (like the red rectangle in the example pictures), but you could assume the border of the maze are all walls.

4. The maze contains at least 2 empty spaces, and both the width and height of the maze won't exceed 100.

Analysis

寻找Shortest Path，一般来说用Dijkstra，或者BFS。不过这里用DFS也可以通过OJ。

Solution

BFS （Queue for next position to be traversed)

Note: For the BFS approach, why is the time complexity of Maze 1 O(m_n) and for Maze 2 its O(m_n*max(m, n)).

Hint: We are NOT exploring each cell EXACTLY ONCE in this problem.

class Solution {
    public int shortestDistance(int[][] maze, int[] start, int[] destination) {
        int m = maze.length;
        int n = maze[0].length;

        int[][] dirs = {{1, 0}, {0, 1}, {0, -1}, {-1, 0}};
        int[][] distance = new int[m][n];

        for (int[] row: distance) {
            Arrays.fill(row, Integer.MAX_VALUE);
        }

        Deque<int[]> queue = new ArrayDeque<>();

        queue.offer(start);
        distance[start[0]][start[1]] = 0;

        while (!queue.isEmpty()) {
            int[] p = queue.poll();

            for (int[] dir: dirs) {
                int nrow = p[0];
                int ncol = p[1];
                int count = 0;

                while (canRoll(maze, nrow + dir[0], ncol + dir[1])) {
                    nrow += dir[0];
                    ncol += dir[1];
                    count++;
                }

                // Update the distance[][], and use to check if a position is visited or not
                if (distance[p[0]][p[1]] + count < distance[nrow][ncol]) {
                    distance[nrow][ncol] = distance[p[0]][p[1]] + count;
                    queue.offer(new int[] {nrow, ncol});
                }

            }
        }
        return distance[destination[0]][destination[1]] == Integer.MAX_VALUE ? 
                    -1 : 
                    distance[destination[0]][destination[1]];
    }

    private boolean canRoll(int[][] maze, int nrow, int ncol) {
        int m = maze.length;
        int n = maze[0].length;
        return nrow >= 0 && nrow < m && ncol >= 0 && ncol < n && maze[nrow][ncol] == 0;
    }
}

• Time complexity : O(m∗n∗max(m,n)). Complete traversal of maze will be done in the worst case. Here, m and n refers to the number of rows and columns of the maze. Further, for every current node chosen, we can travel up to a maximum depth of max(m,n) in any direction.

• Space complexity :O(mn).queuequeuesize can grow up to m*nin the worst case.

Dijkstra (PriorityQueue - min heap for selecting next node to traverse)

public class Solution {
    public int shortestDistance(int[][] maze, int[] start, int[] dest) {
        int[][] distance = new int[maze.length][maze[0].length];
        for (int[] row: distance)
            Arrays.fill(row, Integer.MAX_VALUE);
        distance[start[0]][start[1]] = 0;
        dijkstra(maze, start, distance);
        return distance[dest[0]][dest[1]] == Integer.MAX_VALUE ? -1 : distance[dest[0]][dest[1]];
    }

    public void dijkstra(int[][] maze, int[] start, int[][] distance) {
        int[][] dirs={{0,1},{0,-1},{-1,0},{1,0}};
        PriorityQueue < int[] > queue = new PriorityQueue < > ((a, b) -> a[2] - b[2]);
        queue.offer(new int[]{start[0],start[1],0});
        while (!queue.isEmpty()) {
            int[] s = queue.poll();
            if(distance[s[0]][s[1]] < s[2])
                continue;
            for (int[] dir: dirs) {
                int x = s[0] + dir[0];
                int y = s[1] + dir[1];
                int count = 0;
                while (x >= 0 && y >= 0 && x < maze.length && y < maze[0].length && maze[x][y] == 0) {
                    x += dir[0];
                    y += dir[1];
                    count++;
                }
                if (distance[s[0]][s[1]] + count < distance[x - dir[0]][y - dir[1]]) {
                    distance[x - dir[0]][y - dir[1]] = distance[s[0]][s[1]] + count;
                    queue.offer(new int[]{x - dir[0], y - dir[1], distance[x - dir[0]][y - dir[1]]});
                }
            }
        }
    }
}

DFS

public class Solution {
    public int shortestDistance(int[][] maze, int[] start, int[] dest) {
        int[][] distance = new int[maze.length][maze[0].length];
        for (int[] row: distance)
            Arrays.fill(row, Integer.MAX_VALUE);
        distance[start[0]][start[1]] = 0;
        dfs(maze, start, distance);
        return distance[dest[0]][dest[1]] == Integer.MAX_VALUE ? -1 : distance[dest[0]][dest[1]];
    }

    public void dfs(int[][] maze, int[] start, int[][] distance) {
        int[][] dirs={{0,1}, {0,-1}, {-1,0}, {1,0}};
        for (int[] dir: dirs) {
            int x = start[0] + dir[0];
            int y = start[1] + dir[1];
            int count = 0;
            while (x >= 0 && y >= 0 && x < maze.length && y < maze[0].length && maze[x][y] == 0) {
                x += dir[0];
                y += dir[1];
                count++;
            }
            if (distance[start[0]][start[1]] + count < distance[x - dir[0]][y - dir[1]]) {
                distance[x - dir[0]][y - dir[1]] = distance[start[0]][start[1]] + count;
                dfs(maze, new int[]{x - dir[0],y - dir[1]}, distance);
            }
        }
    }
}

Reference

https://leetcode.com/problems/the-maze-ii/solution/