505. The Maze II - Explanation

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 the shortest distance for the ball to stop at the destination. If the ball cannot stop at destination, return -1.

The distance is the number of empty spaces traveled by the ball from the start position (excluded) to the destination (included).

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: 12

Explanation: One possible way is : left -> down -> left -> down -> right -> down -> right.
The length of the path is 1 + 1 + 3 + 1 + 2 + 2 + 2 = 12.

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: -1

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: -1

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.

Company Tags

1. Depth First Search

Java

class Solution {
    public int shortestDistance(int[][] maze, int[] start, int[] dest) {
        int[][] distance = new int[maze.length][maze[0].length];
        for (int[] row: distance)
            Arrays.fill(row, Integer.MAX_VALUE);
        distance[start[0]][start[1]] = 0;
        dfs(maze, start, distance);
        return distance[dest[0]][dest[1]] == Integer.MAX_VALUE ? -1 : distance[dest[0]][dest[1]];
    }

    public void dfs(int[][] maze, int[] start, int[][] distance) {
        int[][] dirs={{0,1}, {0,-1}, {-1,0}, {1,0}};
        for (int[] dir: dirs) {
            int x = start[0] + dir[0];
            int y = start[1] + dir[1];
            int count = 0;
            while (x >= 0 && y >= 0 && x < maze.length && y < maze[0].length && maze[x][y] == 0) {
                x += dir[0];
                y += dir[1];
                count++;
            }
            if (distance[start[0]][start[1]] + count < distance[x - dir[0]][y - dir[1]]) {
                distance[x - dir[0]][y - dir[1]] = distance[start[0]][start[1]] + count;
                dfs(maze, new int[]{x - dir[0],y - dir[1]}, distance);
            }
        }
    }
}

Time & Space Complexity

Time complexity: $O(m \cdot n \cdot max(m,n))$
Space complexity: $O(m \cdot n)$

Where $m$ and $n$ are the number of rows and columns in maze.

2. Breadth First Search

Java

class Solution {
    public int shortestDistance(int[][] maze, int[] start, int[] dest) {
        int[][] distance = new int[maze.length][maze[0].length];
        for (int[] row: distance)
            Arrays.fill(row, Integer.MAX_VALUE);

        distance[start[0]][start[1]] = 0;
        int[][] dirs={{0, 1} ,{0, -1}, {-1, 0}, {1, 0}};
        Queue < int[] > queue = new LinkedList < > ();
        queue.add(start);
        while (!queue.isEmpty()) {
            int[] s = queue.remove();
            for (int[] dir: dirs) {
                int x = s[0] + dir[0];
                int y = s[1] + dir[1];
                int count = 0;
                while (x >= 0 && y >= 0 && x < maze.length && y < maze[0].length && maze[x][y] == 0) {
                    x += dir[0];
                    y += dir[1];
                    count++;
                }

                if (distance[s[0]][s[1]] + count < distance[x - dir[0]][y - dir[1]]) {
                    distance[x - dir[0]][y - dir[1]] = distance[s[0]][s[1]] + count;
                    queue.add(new int[] {x - dir[0], y - dir[1]});
                }
            }
        }
        return distance[dest[0]][dest[1]] == Integer.MAX_VALUE ? -1 : distance[dest[0]][dest[1]];
    }
}

Time & Space Complexity

Time complexity: $O(m \cdot n \cdot (m + n))$
Space complexity: $O(m \cdot n)$

Where $m$ and $n$ are the number of rows and columns in maze.

3. Dijkstra's Algorithm

Java

class Solution {
    public int shortestDistance(int[][] maze, int[] start, int[] dest) {
        int[][] distance = new int[maze.length][maze[0].length];
        boolean[][] visited = new boolean[maze.length][maze[0].length];
        for (int[] row: distance)
            Arrays.fill(row, Integer.MAX_VALUE);

        distance[start[0]][start[1]] = 0;
        dijkstra(maze, distance, visited);
        return distance[dest[0]][dest[1]] == Integer.MAX_VALUE ? -1 : distance[dest[0]][dest[1]];
    }

    public int[] minDistance(int[][] distance, boolean[][] visited) {
        int[] min={-1,-1};
        int min_val = Integer.MAX_VALUE;

        for (int i = 0; i < distance.length; i++) {
            for (int j = 0; j < distance[0].length; j++) {
                if (!visited[i][j] && distance[i][j] < min_val) {
                    min = new int[] {i, j};
                    min_val = distance[i][j];
                }
            }
        }

        return min;
    }

    public void dijkstra(int[][] maze, int[][] distance, boolean[][] visited) {
        int[][] dirs={{0,1},{0,-1},{-1,0},{1,0}};
        while (true) {
            int[] s = minDistance(distance, visited);
            if (s[0] < 0)
                break;
            visited[s[0]][s[1]] = true;
            for (int[] dir: dirs) {
                int x = s[0] + dir[0];
                int y = s[1] + dir[1];
                int count = 0;
                while (x >= 0 && y >= 0 && x < maze.length && y < maze[0].length && maze[x][y] == 0) {
                    x += dir[0];
                    y += dir[1];
                    count++;
                }

                if (distance[s[0]][s[1]] + count < distance[x - dir[0]][y - dir[1]]) {
                    distance[x - dir[0]][y - dir[1]] = distance[s[0]][s[1]] + count;
                }
            }
        }
    }
}

Time & Space Complexity

Time complexity: $O((mn)^2)$
Space complexity: $O(mn)$

Where $m$ and $n$ are the number of rows and columns in maze.

4. Dijkstra's Algorithm and Priority Queue

class Solution:
    def shortestDistance(self, maze: List[List[int]], start: List[int], destination: List[int]) -> int:
        distance = [[float('inf')] * len(maze[0]) for _ in range(len(maze))]
        distance[start[0]][start[1]] = 0
        self.dijkstra(maze, start, distance)
        return -1 if distance[destination[0]][destination[1]] == float('inf') else distance[destination[0]][destination[1]]
    
    def dijkstra(self, maze: List[List[int]], start: List[int], distance: List[List[int]]) -> None:
        dirs = [[0, 1], [0, -1], [-1, 0], [1, 0]]
        queue = []
        heapq.heappush(queue, (0, start[0], start[1]))  # (distance, x, y)
        
        while queue:
            dist, sx, sy = heapq.heappop(queue)
            
            if distance[sx][sy] < dist:
                continue
            
            for dx, dy in dirs:
                x, y = sx + dx, sy + dy
                count = 0
                
                while 0 <= x < len(maze) and 0 <= y < len(maze[0]) and maze[x][y] == 0:
                    x += dx
                    y += dy
                    count += 1
                
                if distance[sx][sy] + count < distance[x - dx][y - dy]:
                    distance[x - dx][y - dy] = distance[sx][sy] + count
                    heapq.heappush(queue, (distance[x - dx][y - dy], x - dx, y - dy))

Time & Space Complexity

Time complexity: $O(mn \cdot \log(mn))$
Space complexity: $O(m \cdot n)$

Where $m$ and $n$ are the number of rows and columns in maze.