the-maze-ii.py

"""

[LeetCode] 505. The Maze II

# https://www.cnblogs.com/grandyang/p/6725380.html
 

There is a ball in a maze with empty spaces and walls. The ball can go through 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 ball's start position, the destination and the maze, find the shortest distance for the ball to stop at the destination. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the destination (included). If the ball cannot stop at the destination, return -1.

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: 12
Explanation: One shortest way is : left -> down -> left -> down -> right -> down -> right.
             The total distance is 1 + 1 + 3 + 1 + 2 + 2 + 2 = 12.


 

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: -1
Explanation: There is no way for the ball to stop at the destination.


 

Note:

There is only one ball and one destination in the maze.
Both the ball and the destination exist on an empty space, and they will not be at the same position initially.
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.
The maze contains at least 2 empty spaces, and both the width and height of the maze won't exceed 100.
 

"""

# V0 

# V1
# https://www.jiuzhang.com/solution/the-maze-ii/#tag-other-lang-python
from collections import deque
class Solution:
    def shortestDistance(self, maze, start, destination):
        queue = deque([(start[0], start[1], 0)])
        visited = {(start[0], start[1])}
        directions = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        minDist = -1
        while queue:
            size = len(queue)
            for _ in range(size):
                x, y, front = queue.popleft()
                if x == destination[0] and y == destination[1]:
                    minDist = front if minDist == -1 else min(minDist, front)

                for dx, dy in directions:
                    nextX, nextY = x + dx, y + dy
                    count = 0
                    while self.isValid(maze, nextX, nextY):
                        nextX += dx
                        nextY += dy
                        count += 1
                    nextX -= dx
                    nextY -= dy
                    if (nextX, nextY) not in visited:
                        queue.append((nextX, nextY, front + count))
                        visited.add((nextX, nextY))
        return minDist

    def isValid(self, maze, x, y):
        rows, cols = len(maze), len(maze[0])
        return 0 <= x < rows and 0 <= y < cols and maze[x][y] == 0

# V1'
# http://bookshadow.com/weblog/2017/01/29/leetcode-the-maze-ii/
# IDEA : BFS 
import collections
class Solution(object):
    def findShortestWay(self, maze, ball, hole):
        """
        :type maze: List[List[int]]
        :type ball: List[int]
        :type hole: List[int]
        :rtype: str
        """
        ball, hole = tuple(ball), tuple(hole)
        dmap = collections.defaultdict(lambda: collections.defaultdict(int))
        w, h = len(maze), len(maze[0])
        for dir in 'dlru': dmap[hole][dir] = hole
        for x in range(w):
            for y in range(h):
                if maze[x][y] or (x, y) == hole: continue
                dmap[(x, y)]['u'] = dmap[(x - 1, y)]['u'] if x > 0 and dmap[(x - 1, y)]['u'] else (x, y)
                dmap[(x, y)]['l'] = dmap[(x, y - 1)]['l'] if y > 0 and dmap[(x, y - 1)]['l'] else (x, y)
        for x in range(w - 1, -1, -1):
            for y in range(h - 1, -1, -1):
                if maze[x][y] or (x, y) == hole: continue
                dmap[(x, y)]['d'] = dmap[(x + 1, y)]['d'] if x < w - 1 and dmap[(x + 1, y)]['d'] else (x, y)
                dmap[(x, y)]['r'] = dmap[(x, y + 1)]['r'] if y < h - 1 and dmap[(x, y + 1)]['r'] else (x, y)
        bmap = {ball : (0, '')}
        distance = lambda pa, pb: abs(pa[0] - pb[0]) + abs(pa[1] - pb[1])
        queue = collections.deque([(ball, 0, '')])
        while queue:
            front, dist, path = queue.popleft()
            for dir in 'dlru':
                if dir not in dmap[front]: continue
                np = dmap[front][dir]
                ndist = dist + distance(front, np)
                npath = path + dir
                if np not in bmap or (ndist, npath) < bmap[np]:
                    bmap[np] = (ndist, npath)
                    queue.append((np, ndist, npath))
        return bmap[hole][1] if hole in bmap else 'impossible'

# V1''
# https://leetcode.com/articles/the-maze-ii/?page=4
# IDEA : DFS
# JAVA 
# public 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);
#             }
#         }
#     }
# }

# V1'''
# http://bookshadow.com/weblog/2017/01/29/leetcode-the-maze-ii/
# IDEA : Dijkstra ALGORITHM
import collections
class Solution(object):
    def findShortestWay(self, maze, ball, hole):
        """
        :type maze: List[List[int]]
        :type ball: List[int]
        :type hole: List[int]
        :rtype: str
        """
        ball, hole = tuple(ball), tuple(hole)
        dmap = collections.defaultdict(lambda: collections.defaultdict(int))
        w, h = len(maze), len(maze[0])
        for dir in 'dlru': dmap[hole][dir] = hole
        for x in range(w):
            for y in range(h):
                if maze[x][y] or (x, y) == hole: continue
                dmap[(x, y)]['u'] = dmap[(x - 1, y)]['u'] if x > 0 and dmap[(x - 1, y)]['u'] else (x, y)
                dmap[(x, y)]['l'] = dmap[(x, y - 1)]['l'] if y > 0 and dmap[(x, y - 1)]['l'] else (x, y)
        for x in range(w - 1, -1, -1):
            for y in range(h - 1, -1, -1):
                if maze[x][y] or (x, y) == hole: continue
                dmap[(x, y)]['d'] = dmap[(x + 1, y)]['d'] if x < w - 1 and dmap[(x + 1, y)]['d'] else (x, y)
                dmap[(x, y)]['r'] = dmap[(x, y + 1)]['r'] if y < h - 1 and dmap[(x, y + 1)]['r'] else (x, y)
        bmap = {ball : (0, '', ball)}
        vset = set()
        distance = lambda pa, pb: abs(pa[0] - pb[0]) + abs(pa[1] - pb[1])
        while bmap:
            dist, path, p = min(bmap.values())
            if p == hole: return path
            del bmap[p]
            vset.add(p)
            for dir in 'dlru':
                if dir not in dmap[p]: continue
                np = dmap[p][dir]
                ndist = dist + distance(p, np)
                npath = path + dir
                if np not in vset and (np not in bmap or (ndist, npath, np) < bmap[np]):
                    bmap[np] = (ndist, npath, np)
        return 'impossible'
        
# V1''''
# https://www.jiuzhang.com/solution/the-maze-ii/#tag-other-lang-python
# IDEA : BFS 
import collections
def shortestDistance(self, maze, start, destination):
        # write your code here
        D=[[-1,0],[1,0],[0,1],[0,-1]]
        n,m=len(maze),len(maze[0])
        q=collections.deque([[start[0],start[1],0]])     
        result=float('inf')
        while q:
            x,y,l=q.popleft()
            maze[x][y]=2
            if x==destination[0] and y==destination[1]:
                result=min(l,result)

            tmp=l
            for i,j in D:
                row=x+i
                col=y+j
                l+=1
                while 0<=row<n and 0<=col<m and maze[row][col]!=1:
                    row+=i
                    col+=j
                    l+=1
                
                row-=i
                col-=j
                l-=1
                if maze[row][col]==0:
                    q.append([row,col,l])
                
                l=tmp

        if result==float('inf'):
            return -1
        return result

# V1'''''
# IDEA : DFS
# JAVA
# class Solution {
# public:
#     vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
#     int shortestDistance(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
#         int m = maze.size(), n = maze[0].size();
#         vector<vector<int>> dists(m, vector<int>(n, INT_MAX));
#         dists[start[0]][start[1]] = 0;
#         helper(maze, start[0], start[1], destination, dists);
#         int res = dists[destination[0]][destination[1]];
#         return (res == INT_MAX) ? -1 : res;
#     }
#     void helper(vector<vector<int>>& maze, int i, int j, vector<int>& destination, vector<vector<int>>& dists) {
#         if (i == destination[0] && j == destination[1]) return;
#         int m = maze.size(), n = maze[0].size();
#         for (auto d : dirs) {
#             int x = i, y = j, dist = dists[x][y];
#             while (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == 0) {
#                 x += d[0];
#                 y += d[1];
#                 ++dist;
#             }
#             x -= d[0];
#             y -= d[1];
#             --dist;
#             if (dists[x][y] > dist) {
#                 dists[x][y] = dist;
#                 helper(maze, x, y, destination, dists);
#             }
#         }
#     }
# };

# V2 
# Time:  O(max(r, c) * wlogw)
# Space: O(w)
import heapq
class Solution(object):
    def shortestDistance(self, maze, start, destination):
        """
        :type maze: List[List[int]]
        :type start: List[int]
        :type destination: List[int]
        :rtype: int
        """
        start, destination = tuple(start), tuple(destination)

        def neighbors(maze, node):
            for dir in [(-1, 0), (0, 1), (0, -1), (1, 0)]:
                cur_node, dist = list(node), 0
                while 0 <= cur_node[0]+dir[0] < len(maze) and \
                      0 <= cur_node[1]+dir[1] < len(maze[0]) and \
                      not maze[cur_node[0]+dir[0]][cur_node[1]+dir[1]]:
                    cur_node[0] += dir[0]
                    cur_node[1] += dir[1]
                    dist += 1
                yield dist, tuple(cur_node)

        heap = [(0, start)]
        visited = set()
        while heap:
            dist, node = heapq.heappop(heap)
            if node in visited: continue
            if node == destination:
                return dist
            visited.add(node)
            for neighbor_dist, neighbor in neighbors(maze, node):
                heapq.heappush(heap, (dist+neighbor_dist, neighbor))

        return -1