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490. The Maze πŸ”’

Description

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 = [startrow, startcol] and destination = [destinationrow, destinationcol], 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).

 

Example 1:

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.

Example 2:

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.

Example 3:

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

 

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 <= startrow, destinationrow < m
  • 0 <= startcol, destinationcol < 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.

Solutions

Solution 1: DFS

Thinking

The ball rolls until it hits a wall; we ask whether it can stop on the destination. Walking cell by cell confuses β€œpassing through” with β€œstopping”.

DFS: from a stop, roll in each direction to a wall or border and recurse on that stop. A visited grid marks stops only. The destination is reachable once it is marked.

The inner \(\textit{while}\) rolls without marking cells on the way; the state space is stops, not every empty cell.

Roll in four directions until hitting a wall, and DFS every stoppable cell from the start.

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class Solution:
    def hasPath(
        self, maze: List[List[int]], start: List[int], destination: List[int]
    ) -> bool:
        def dfs(i, j):
            if vis[i][j]:
                return
            vis[i][j] = True
            if [i, j] == destination:
                return
            for a, b in [[0, -1], [0, 1], [1, 0], [-1, 0]]:
                x, y = i, j
                while 0 <= x + a < m and 0 <= y + b < n and maze[x + a][y + b] == 0:
                    x, y = x + a, y + b
                dfs(x, y)

        m, n = len(maze), len(maze[0])
        vis = [[False] * n for _ in range(m)]
        dfs(start[0], start[1])
        return vis[destination[0]][destination[1]]

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class Solution {
    private boolean[][] vis;
    private int[][] maze;
    private int[] d;
    private int m;
    private int n;

    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
        m = maze.length;
        n = maze[0].length;
        d = destination;
        this.maze = maze;
        vis = new boolean[m][n];
        dfs(start[0], start[1]);
        return vis[d[0]][d[1]];
    }

    private void dfs(int i, int j) {
        if (vis[i][j]) {
            return;
        }
        vis[i][j] = true;
        if (i == d[0] && j == d[1]) {
            return;
        }
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i, y = j;
            int a = dirs[k], b = dirs[k + 1];
            while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && maze[x + a][y + b] == 0) {
                x += a;
                y += b;
            }
            dfs(x, y);
        }
    }
}

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class Solution {
public:
    vector<vector<int>> maze;
    vector<vector<bool>> vis;
    vector<int> d;
    int m;
    int n;

    bool hasPath(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
        m = maze.size();
        n = maze[0].size();
        d = destination;
        vis.resize(m, vector<bool>(n, false));
        this->maze = maze;
        dfs(start[0], start[1]);
        return vis[d[0]][d[1]];
    }

    void dfs(int i, int j) {
        if (vis[i][j]) return;
        vis[i][j] = true;
        if (i == d[0] && j == d[1]) return;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i, y = j;
            int a = dirs[k], b = dirs[k + 1];
            while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && maze[x + a][y + b] == 0) {
                x += a;
                y += b;
            }
            dfs(x, y);
        }
    }
};

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func hasPath(maze [][]int, start []int, destination []int) bool {
    m, n := len(maze), len(maze[0])
    vis := make([][]bool, m)
    for i := range vis {
        vis[i] = make([]bool, n)
    }
    var dfs func(i, j int)
    dfs = func(i, j int) {
        if vis[i][j] {
            return
        }
        vis[i][j] = true
        if i == destination[0] && j == destination[1] {
            return
        }
        dirs := []int{-1, 0, 1, 0, -1}
        for k := 0; k < 4; k++ {
            x, y := i, j
            a, b := dirs[k], dirs[k+1]
            for x+a >= 0 && x+a < m && y+b >= 0 && y+b < n && maze[x+a][y+b] == 0 {
                x += a
                y += b
            }
            dfs(x, y)
        }
    }
    dfs(start[0], start[1])
    return vis[destination[0]][destination[1]]
}

Solution 2: BFS

Thinking

Solution 1 is recursive. The same rolling rule works with a queue: enqueue stops, and return as soon as the destination is reached. No call-stack depth, which is enough for a yes/no reachability query.

Roll in four directions and BFS the stoppable cells until the destination is reached.

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class Solution:
    def hasPath(
        self, maze: List[List[int]], start: List[int], destination: List[int]
    ) -> bool:
        m, n = len(maze), len(maze[0])
        q = deque([start])
        rs, cs = start
        vis = {(rs, cs)}
        while q:
            i, j = q.popleft()
            for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
                x, y = i, j
                while 0 <= x + a < m and 0 <= y + b < n and maze[x + a][y + b] == 0:
                    x, y = x + a, y + b
                if [x, y] == destination:
                    return True
                if (x, y) not in vis:
                    vis.add((x, y))
                    q.append((x, y))
        return False

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class Solution {
    public boolean hasPath(int[][] maze, int[] start, int[] destination) {
        int m = maze.length;
        int n = maze[0].length;
        boolean[][] vis = new boolean[m][n];
        vis[start[0]][start[1]] = true;
        Deque<int[]> q = new LinkedList<>();
        q.offer(start);
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            int[] p = q.poll();
            int i = p[0], j = p[1];
            for (int k = 0; k < 4; ++k) {
                int x = i, y = j;
                int a = dirs[k], b = dirs[k + 1];
                while (
                    x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && maze[x + a][y + b] == 0) {
                    x += a;
                    y += b;
                }
                if (x == destination[0] && y == destination[1]) {
                    return true;
                }
                if (!vis[x][y]) {
                    vis[x][y] = true;
                    q.offer(new int[] {x, y});
                }
            }
        }
        return false;
    }
}

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class Solution {
public:
    bool hasPath(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
        int m = maze.size();
        int n = maze[0].size();
        queue<vector<int>> q{{start}};
        vector<vector<bool>> vis(m, vector<bool>(n));
        vis[start[0]][start[1]] = true;
        vector<int> dirs = {-1, 0, 1, 0, -1};
        while (!q.empty()) {
            auto p = q.front();
            q.pop();
            int i = p[0], j = p[1];
            for (int k = 0; k < 4; ++k) {
                int x = i, y = j;
                int a = dirs[k], b = dirs[k + 1];
                while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && maze[x + a][y + b] == 0) {
                    x += a;
                    y += b;
                }
                if (x == destination[0] && y == destination[1]) return 1;
                if (!vis[x][y]) {
                    vis[x][y] = true;
                    q.push({x, y});
                }
            }
        }
        return 0;
    }
};

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func hasPath(maze [][]int, start []int, destination []int) bool {
    m, n := len(maze), len(maze[0])
    vis := make([][]bool, m)
    for i := range vis {
        vis[i] = make([]bool, n)
    }
    vis[start[0]][start[1]] = true
    q := [][]int{start}
    dirs := []int{-1, 0, 1, 0, -1}
    for len(q) > 0 {
        i, j := q[0][0], q[0][1]
        q = q[1:]
        for k := 0; k < 4; k++ {
            x, y := i, j
            a, b := dirs[k], dirs[k+1]
            for x+a >= 0 && x+a < m && y+b >= 0 && y+b < n && maze[x+a][y+b] == 0 {
                x += a
                y += b
            }
            if x == destination[0] && y == destination[1] {
                return true
            }
            if !vis[x][y] {
                vis[x][y] = true
                q = append(q, []int{x, y})
            }
        }
    }
    return false
}

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