LeetCode: Maze

/**
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, determine whether the ball could stop at the destination.

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: true

Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.

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: false

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



Note:

There is only one ball and one destination in the maze.
Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
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.
The maze contains at least 2 empty spaces, and both the width and height of the maze won't exceed 100.
*/

public class Maze{


public boolean hasPath(int[][] maze, int[] start, int[] destination) {
        boolean[][] visited = new boolean[maze.length][maze[0].length];
        return dfs(maze, start, destination, visited);
    }

/**
Sol1:
-use dfs from the start position and traverse on left, right, up, and down positions
-if we reach the destination at the extreme point (the cell near the wall), then we return true
O(mn) time and O(mn) space to track the visited cells
*/
    public boolean dfs(int[][] maze, int[] start, int[] destination, boolean[][] visited) {
        if (visited[start[0]][start[1]])
            return false;
        if (start[0] == destination[0] && start[1] == destination[1])
            return true;
        visited[start[0]][start[1]] = true;
//right increases the col numbers, left decreases the col numbers
//up decreases the row numbers, down increases the row numbers
//start[0] is the row and start[1] is the col
        int r = start[1] + 1, l = start[1] - 1, u = start[0] - 1, d = start[0] + 1;
        while (r < maze[0].length && maze[start[0]][r] == 0) // right
            r++;
        if (dfs(maze, new int[] {start[0], r - 1}, destination, visited))
            return true;
        while (l >= 0 && maze[start[0]][l] == 0) //left
            l--;
        if (dfs(maze, new int[] {start[0], l + 1}, destination, visited))
            return true;
        while (u >= 0 && maze[u][start[1]] == 0) //up
            u--;
        if (dfs(maze, new int[] {u + 1, start[1]}, destination, visited))
            return true;
        while (d < maze.length && maze[d][start[1]] == 0) //down
            d++;
        if (dfs(maze, new int[] {d - 1, start[1]}, destination, visited))
            return true;
        return false;
    }

/**
-use BFS from teh start point
-mark start as visited
-add the left, right, up, and down cells (if not wall) to a queue
-repeat while the queue is not empty
-poll a point from queue and check if we found destination
-get four neighbors for this point/cell
-if the neighbors are not visited, add them to queue
O(mn) time and O(mn) space
*/
public boolean hasPathBFS(int[][] maze, int[] start, int[] destination) {
        boolean[][] visited = new boolean[maze.length][maze[0].length];
        int[][] dirs={{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
        Queue < int[] > queue = new LinkedList < > ();
        queue.add(start);
        visited[start[0]][start[1]] = true;
        while (!queue.isEmpty()) {
            int[] s = queue.remove();
            if (s[0] == destination[0] && s[1] == destination[1])
                return true;
            for (int[] dir: dirs) {
                int x = s[0] + dir[0];
                int y = s[1] + dir[1];
                while (x >= 0 && y >= 0 && x < maze.length && y < maze[0].length && maze[x][y] == 0) {
                    x += dir[0];
                    y += dir[1];
                }
                if (!visited[x - dir[0]][y - dir[1]]) {
                    queue.add(new int[] {x - dir[0], y - dir[1]});
                    visited[x - dir[0]][y - dir[1]] = true;
                }
            }
        }
        return false;
    }
}