The Maze

class Solution {
public:
    bool canReach(vector<vector<int>>& grid, vector<int>& start, vector<int>& end) {
        int rows = grid.size();
        int cols = grid[0].size();
        vector<vector<bool>> visited(rows, vector<bool>(cols, false));
        vector<int> deltaX{0, 1, 0, -1};
        vector<int> deltaY{-1, 0, 1, 0};
    
        queue<vector<int>> queue;
        queue.push(start);
        visited[start[0]][start[1]] = true;
    
        while (!queue.empty()) {
            vector<int> current = queue.front();
            queue.pop();
            if (current[0] == end[0] && current[1] == end[1]) {
                return true;
            }
    
            for (int i = 0; i < 4; i++) {
                int x = current[0];
                int y = current[1];
              
                while (x >= 0 && x < rows && y >= 0 && y < cols && grid[x][y] == 0) {
                    x += deltaX[i];
                    y += deltaY[i];
                }
              
                x -= deltaX[i];
                y -= deltaY[i];
                if (!visited[x][y]) {
                    queue.push({x, y});
                    visited[x][y] = true;
                }
            }
        }
        return false;
    }
};

The Maze problem involves determining whether there is a path from a start point to an endpoint in a 2D grid represented by 0s and 1s, where 0 indicates an open path and 1 indicates a wall.

• You can solve this problem using a Breadth-First Search (BFS) approach:

  • Initialize a queue and add the start position to it. The visited 2D vector keeps track of visited nodes to avoid revisiting them.
  • While the queue is not empty, dequeue the front element. If this element is the end position, return true to indicate that the endpoint is reachable.
  • For each possible move (up, down, left, and right), calculate the next position until hitting a wall or the boundary of the grid. Then move back one step to the last valid position.
  • If this new position has not been visited, mark it as visited, and enqueue it for further exploration.
  • If you exhaust all possibilities and do not find the end position, return false as it is not reachable.

This solution efficiently explores the grid by only visiting each potential path once and backtracking only when necessary. Continue optimizing based on the specific constraints and sizes of the input grid.

class MazeSolver {
    public boolean canReach(int[][] maze, int[] start, int[] end) {
        // Getting maze dimensions
        int rows = maze.length;
        int cols = maze[0].length;
            
        // Visited cells tracker
        boolean[][] visited = new boolean[rows][cols];
            
        // Directions arrays
        int[] deltaX = {0, 1, 0, -1};
        int[] deltaY = {-1, 0, 1, 0};
    
        // Queue for breadth-first search
        Queue<int[]> bfsQueue = new LinkedList<>();
        bfsQueue.offer(start);
        visited[start[0]][start[1]] = true;
    
        while (!bfsQueue.isEmpty()) {
            int[] current = bfsQueue.poll();
            if (current[0] == end[0] && current[1] == end[1]) {
                return true;
            }
            for (int i = 0; i < 4; i++) {
                int x = current[0];
                int y = current[1];
                // Continue in direction until hitting a wall
                while (x >= 0 && x < rows && y >= 0 && y < cols && maze[x][y] == 0) {
                    x += deltaX[i];
                    y += deltaY[i];
                }
                // Backtrack last move
                x -= deltaX[i];
                y -= deltaY[i];
                if (!visited[x][y]) {
                    bfsQueue.offer(new int[]{x, y});
                    visited[x][y] = true;
                }
            }
        }
        return false;
    }
}

The "The Maze" problem is efficiently addressed using the Java programming language in the provided solution. Here's a breakdown on how the code operates and accomplishes the task of finding a path in the maze:

• Maze Representation and Input: The maze is represented as a 2D array where 0 signifies a passable cell and 1 an impassable wall. Inputs to the function include the maze itself and start/end coordinates as arrays.

• Data Structures Used:

  • Visited Array: A 2D boolean array that keeps track of visited cells.
  • BFS Queue: A Queue holding array coordinates facilitating a breadth-first search.

• Algorithm Initialization:

  • The start cell is marked as visited and then added to the BFS queue.

• Breadth-First Search (BFS) Implementation:

  1. Begin a loop over the BFS queue until it's empty, processing each cell.
  2. Check each dequeued cell. If it matches the end location, return true indicating a path exists.
  3. Explore possible movements from the current cell in four directions (up, right, down, left) using arrays deltaX and deltaY.
  4. Upon encountering walls or boundaries, backtrack to the last valid cell.
  5. If this new cell hasn’t been visited, mark it as visited and enqueue it.

• End of Algorithm: If no path is found by the time the queue empties, return false.

This code uses BFS for exploration, which ensures that the shortest path in an unweighted maze is found due to the layer-by-layer nature of BFS. Furthermore, the solution is robust in adhering to boundaries and avoiding revisits, making it efficient and effective for solving large mazes. The use of a queue in conjunction with the visited matrix optimally fosters both memory and performance efficiency.

class Solution:
    def pathExists(self, grid: List[List[int]], source: List[int], target: List[int]) -> bool:
        rows = len(grid)
        cols = len(grid[0])
        visited = [[False] * cols for _ in range(rows)]
        q = deque()
            
        q.append(source)
        visited[source[0]][source[1]] = True
        directionsX = [0, 1, 0, -1]
        directionsY = [-1, 0, 1, 0]
    
        while q:
            current = q.popleft()
            if current[0] == target[0] and current[1] == target[1]:
                return True
    
            for j in range(4):
                x = current[0]
                y = current[1]
                # Continue moving in the direction until hitting a wall or boundary
                while x >= 0 and x < rows and y >= 0 and y < cols and grid[x][y] == 0:
                    x += directionsX[j]
                    y += directionsY[j]
                # Move back to the last valid position
                x -= directionsX[j]
                y -= directionsY[j]
                if not visited[x][y]:
                    q.append([x, y])
                    visited[x][y] = True
        return False

The provided Python solution implements a method to determine if a path exists between a source and target point within a maze represented as a grid. The grid is a 2D list where open paths are denoted by 0 and walls by 1. The solution leverages Breadth-First Search (BFS) to explore the maze.

• Initialize data structures for tracking rows and columns size, visited cells, and BFS queue starting from the source point.
• Continuously explore the maze by dequeuing elements from the queue and checking all possible directions.
• For each direction, move continuously until hitting a wall or boundary, then adjust to the last possible open cell.
• If this cell hasn't been visited, enqueue it for further exploration.
• Return True if the target is reached during the BFS process; otherwise, return False after all possibilities have been exhausted.

The approach ensures that every reachable area within the maze is explored using only valid moves, accounting for maze boundaries and internal walls, and guarantees an efficient search to maintain performance.