The Maze

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

Given the ball'sstart position, thedestinationand themaze, 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:

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.

Solution:

It's a implicit graph search problem. Basically a disguised BFS problem.

Use BFS, a queue and a set. to store possible next steps, and record visited coordinates.

A helper function getNextMoves() to find possible next moves.

Time complexity: O(V+E)

Code:

class Solution(object):
    def hasPath(self, maze, start, destination):
        """
        :type maze: List[List[int]]
        :type start: List[int]
        :type destination: List[int]
        :rtype: bool
        """
        visited = set()
        frontier = [start]
        while frontier:
            nextF = []
            for [row, col] in frontier:
                if [row, col] == destination:
                    return True
                visited.add((row,col)) # when you leave current position, it becomes visited
                moves = self.getNextMoves(maze, [row,col], visited)
                nextF.extend(moves)
            frontier = nextF

        # here it means you have exhaust all possible moves and not finding destination
        return False


    # return the possible next moves in a list, it could return length 0 upto length 4
    def getNextMoves(self, maze, start, visited):
        res = []
        rowNum = len(maze)
        colNum = len(maze[0])

        # roll ball down
        row, col = start
        while row < rowNum and maze[row][col] != 1:
            row += 1
        # it means we go beyond wall or we hit a wall
        # and (row == rowNum or maze[row][col] == 1): # I dont think we need the second half of the check()
        if [row-1,col] != start and (row-1,col) not in visited:
            res.append([row-1, col])

        # roll ball up
        row, col = start
        while row >= 0 and maze[row][col] != 1:
            row -= 1
        if [row+1,col] != start and (row+1,col) not in visited:
            res.append([row+1, col])

        # roll ball left
        row, col = start
        while col >= 0 and maze[row][col] != 1:
            col -= 1
        if [row, col+1] != start and (row,col+1) not in visited:
            res.append([row, col + 1])

        # roll ball right
        row, col = start
        while col < colNum and maze[row][col] != 1:
            col += 1
        if [row, col-1] != start and (row,col-1) not in visited:
            res.append([row, col - 1])

        return res