490. The Maze

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 = [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).

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 <= 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.

Solutions

Solution 1: Depth-First Search

This implementation uses depth-first search. It traverses the relevant values and updates its state as each value is processed. After all required states have been considered, the maintained result is returned.

Solution 2

This implementation uses breadth-first search. It traverses the relevant values and updates its state as each value is processed. The auxiliary container determines which pending state is processed next. After all required states have been considered, the maintained result is returned.

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;
    }
}


// Solution 2
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;
    }
}

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;
    }
};


// Solution 2
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;
    }
};

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


# Solution 2
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

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
}


// Solution 2
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
}

490 - The Maze

490. The Maze

Description

Solutions

Solution 1: Depth-First Search

Solution 2

All Problems

All Solutions

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490. The Maze

Description

Solutions

Solution 1: Depth-First Search

Solution 2

All Problems

All Solutions