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The Maze Iii  Leetcode Solution
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LeetCode: The Maze Iii Leetcode Solution
Difficulty: Hard
Topics: depthfirstsearch string breadthfirstsearch matrix heappriorityqueue array graph
The Maze III problem on LeetCode is an interesting problem involving finding the shortest path from a start point to an end point while taking into consideration the presence of obstacles in the maze. The problem is quite lengthy and challenging, so let's take a closer look.
Problem Statement:
The Maze III problem involves having a maze that has a start point, an end point and several obstacles in between. The task is to find the shortest path from the start point to the end point, considering that the ball moves in a certain direction until it hits a wall. The maze is described using a matrix, where each matrix element represents either a path, an obstacle or the start/end points.
Additionally, the ball can move in four directions namely 'u' (up), 'd' (down), 'l' (left), and 'r' (right). Each move takes up one unit of time and the total time taken is considered while calculating the shortest path.
Function Signature:
The function signature for this problem is as follows:
def findShortestWay(self, maze: List[List[int]], ball: List[int], hole: List[int]) > str:
Where,
'findShortestWay' is the function name.
'maze' is a List[List[int]] representation of the maze.
'ball' is a List[int] containing the starting position of the ball in the maze.
'hole' is a List[int] containing the hole position in the maze.
The function returns a str type value, which represents the shortest path from the start to the end, or the string 'impossible' if it is not possible to reach the end point.
Problem Solution Approach:
This is a classic graph theory problem. The idea is to consider the maze as a graph with each cell being a node, and the edges represent the movement from one cell to another. The goal is to find the shortest path from the starting point to the end point by performing a graph traversal.
This can be achieved efficiently using Dijkstra's shortest path algorithm. However, this algorithm assumes that all edge weights are positive. In our case, each movement to a new cell takes up one unit of time, which can be considered as the weight of the edge. Therefore, the edge's weight is a positive integer. But we have walls in the maze, and moving through them is not possible, and hence should not be considered while traversing the graph.
We can model the problem using a modified BFS (BreadthFirst Search) algorithm with a priority queue. This will ensure that we always consider the node with the lowest time first. The algorithm will work as follows:

Define a priority queue and add ball (starting) node to the queue with priority set as 0 (the start).

Define a hashmap to keep track of the visited cells and updating the value with the minimum time to reach that cell.

While the priority queue is not empty:
a. Pop the cell with the highest priority (lowest time) from the queue.
b. If the popped cell is the hole (the end):
i. Append the path to the result. ii. If the result path is shorter than any previously found paths then update the shortest path. iii. Continue with the loop until the priority queue is empty.
c. For each of the four possible directions:
i. Check the distance that can be travelled until the ball hits a wall. It can either be a wall or the hole itself. In this case, the queue’s priority is updated to the new minimum time to reach this cell. ii. If not already visited, add the cell to the priority queue with its priority set as the new time to reach this cell.

If the end position is not reached at the end of the loop, we return "impossible".
This whole algorithm can be implemented in Python as follows:
Code Implementation:
import heapq
from collections import defaultdict
class Solution:
def findShortestWay(self, maze, ball, hole):
heap = [(0, "", ball[0], ball[1])]
visited = defaultdict(lambda: float('inf'))
while heap:
moves, path, i, j = heapq.heappop(heap)
if visited[(i, j)] <= moves:
continue
visited[(i, j)] = moves
if [i, j] == hole:
return path
for dirn in ['u', 'd', 'l', 'r']:
ii, jj = i, j
while 0 <= ii + (dirn=='d')  (dirn=='u') < len(maze) and \
0 <= jj + (dirn=='r')  (dirn=='l') < len(maze[0]) and \
maze[ii + (dirn=='d')  (dirn=='u')][jj + (dirn=='r')  (dirn=='l')] == 0 and \
[ii, jj] != hole:
ii = ii + (dirn=='d')  (dirn=='u')
jj = jj + (dirn=='r')  (dirn=='l')
heapq.heappush(heap, (moves+iii+jjj, path+dirn, ii, jj))
return "impossible"
Time Complexity Analysis:
The time complexity of the BFS algorithm is linear in the number of nodes in the graph, which is O(V + E). In our case, V = maze size (n x m) and E ≤ 4V (since each node has at most 4 edges). Therefore, the time complexity of the algorithm is O(nm * log(nm)), where log is for the priority queue's push and pop operations.
Space Complexity Analysis:
The space complexity is also linear in the number of nodes, which is O(V + E). In our case, this is O(nm) for both searching and visited sets, and O(V) for the priority queue. Therefore, the space complexity of the algorithm is O(nm).
Conclusion:
The Maze III problem on LeetCode is an interesting problem that requires traversing a graph, and finding the shortest path from a starting point to an end point, considering obstacles in between. By applying a modified BFS algorithm using a priority queue, the shortest path can be achieved efficiently, and hence it is a standard problem of BFS algorithms.
The Maze Iii Solution Code
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