Escape A Large Maze - Leetcode Solution

The problem "Escape A Large Maze" on LeetCode is a graph problem that requires us to find if we can escape a large maze or not.

Problem Statement: We are given a maze represented as an MxN grid. Each grid cell is either blocked or open. We also have an initial position and a destination position. We can move one step at a time in any of the four directions - up, down, left, or right. We need to find if we can reach the destination from the initial position, given that the maze is so large that it may take a very long time to reach the destination.

Approach: Since the maze is so large, we cannot afford to generate the graph explicitly by creating nodes and edges. We need to find an efficient way to represent the graph so that we can perform graph traversal algorithms to find if we can reach the destination or not.

One approach is to use Breadth First Search (BFS) to find the shortest path from the initial position to the destination position. However, since the maze is so large, BFS may take a very long time to find the answer. To optimize the solution, we need to find a way to minimize the search space.

We can take advantage of the fact that the maze is so large and block some of the paths that are unlikely to lead to the destination. We can do this by setting a limit on how far we can go before we consider that we are going in circles, and we cannot reach the destination.

We can use a HashSet to keep track of the visited nodes so that we don't waste time revisiting them. We can also check if the destination is reachable by checking if the distance between the initial and the destination nodes is less than the limit that we set.

Algorithm:

1. Initialize a HashSet to keep track of the visited nodes.
2. Initialize a queue to keep track of the nodes that we need to visit next.
3. Add the initial position to the queue and the visited set.
4. Set a limit L on how far we can go before considering that we are going in circles.
5. While the queue is not empty, do the following: a. Remove the node at the head of the queue. b. Check if this is the destination node. If so, return true. c. If we have exceeded the limit L, return false. d. For each neighboring node that is open and not visited, add it to the queue and mark it as visited.
6. If we have not found the destination node, return false.

Implementation: Here is the Python implementation of the above algorithm:

def isEscapePossible(blocked: List[List[int]], source: List[int], target: List[int]) -> bool: # Initialize a hash set to keep track of visited nodes visited = set() # Initialize a queue to keep track of the nodes that we need to visit next queue = [] # Add the initial position to the queue and the visited set queue.append(tuple(source)) visited.add(tuple(source)) # Set a limit on how far we can go before considering that we are going in circles limit = len(blocked) * len(blocked[0])

while queue:
    # Remove the node at the head of the queue
    curr = queue.pop(0)
    # Check if this is the destination node
    if curr == tuple(target):
        return True
    # If we have exceeded the limit, return false
    if len(visited) >= limit:
        return True
    # Add neighboring nodes that are open and not visited to the queue and mark them as visited
    for neighbor in [(0,1),(1,0),(0,-1),(-1,0)]:
        x, y = curr[0]+neighbor[0], curr[1]+neighbor[1]
        if 0 <= x < len(blocked) and 0 <= y < len(blocked[0]) and not blocked[x][y] and (x,y) not in visited:
            queue.append((x,y))
            visited.add((x,y))
return False

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