The Maze Python Solution with Tests

Tested Python solution for LeetCode 490 with 12 pytest cases. Generate a practice environment with lcpy.

LeetCode 490, Medium. Topics: Depth-First Search, Breadth-First Search, Array, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 12 parametrized pytest cases, and a playground notebook:

lcpy gen -n 490   # by problem number
lcpy gen -s the_maze   # by problem name

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Problem

There is a ball in a maze with empty spaces (represented as 0) and walls (represented as 1). The ball can go through the 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 m x n maze, the ball’s start position and the destination, where start = [start_row, start_col] and destination = [destination_row, destination_col], return true if the ball can stop at the destination, otherwise return false.

You may assume that the borders of the maze are all walls (see examples).

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Examples

Input: maze = [[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]], start = [0,4], destination = [4,4]
Output: true
Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.

Input: maze = [[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]], start = [0,4], destination = [3,2]
Output: false
Explanation: There is no way for the ball to stop at the destination. Notice that you can pass through the destination but you cannot stop there.

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]], start = [4,3], destination = [0,1]
Output: false

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Constraints

• m == maze.length
• n == maze[i].length
• 1 <= m, n <= 100
• maze[i][j] is 0 or 1.
• start.length == 2
• destination.length == 2
• 0 <= start_row, destination_row < m
• 0 <= start_col, destination_col < n
• Both the ball and the destination exist in an empty space, and they will not be in the same position initially.
• The maze contains at least 2 empty spaces.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(m * n * max(m, n))
    # Space: O(m * n)
    def has_path(self, maze: list[list[int]], start: list[int], destination: list[int]) -> bool:
        m, n = len(maze), len(maze[0])
        dirs = ((-1, 0), (1, 0), (0, -1), (0, 1))
        queue = [tuple(start)]
        seen = {tuple(start)}
        while queue:
            r, c = queue.pop(0)
            if [r, c] == destination:
                return True
            for dr, dc in dirs:
                nr, nc = r, c
                while 0 <= nr + dr < m and 0 <= nc + dc < n and maze[nr + dr][nc + dc] == 0:
                    nr += dr
                    nc += dc
                if (nr, nc) not in seen:
                    seen.add((nr, nc))
                    queue.append((nr, nc))
        return False

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Complexity

Time	Space
O(m * n * max(m, n))	O(m * n)

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Tags

NeetCode All.