Escape a Large Maze

Hard

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ArraysGraphs

There is a 1 million by 1 million grid on an XY-plane, and the coordinates of each grid square are (x, y).

We start at the source = [s_x, s_y] square and want to reach the target = [t_x, t_y] square. There is also an array of blocked squares, where each blocked[i] = [x_i, y_i] represents a blocked square with coordinates (x_i, y_i).

Each move, we can walk one square north, east, south, or west if the square is not in the array of blocked squares. We are also not allowed to walk outside of the grid.

Return true if and only if it is possible to reach the target square from the source square through a sequence of valid moves.

Example 1:

Input: blocked = [[0,1],[1,0]], source = [0,0], target = [0,2]
Output: false
Explanation: The target square is inaccessible starting from the source square because we cannot move.
We cannot move north or east because those squares are blocked.
We cannot move south or west because we cannot go outside of the grid.

Example 2:

Input: blocked = [], source = [0,0], target = [999999,999999]
Output: true
Explanation: Because there are no blocked cells, it is possible to reach the target square.

Constraints:

0 <= blocked.length <= 200
blocked[i].length == 2
0 <= x_i, y_i < 10⁶
source.length == target.length == 2
0 <= s_x, s_y, t_x, t_y < 10⁶
source != target
It is guaranteed that source and target are not blocked.

Try coding it in LeetCode

Escape a Large Maze

Hard

UiPath

0 views

Topics:

ArraysGraphs

There is a 1 million by 1 million grid on an XY-plane, and the coordinates of each grid square are (x, y).

We start at the source = [s_x, s_y] square and want to reach the target = [t_x, t_y] square. There is also an array of blocked squares, where each blocked[i] = [x_i, y_i] represents a blocked square with coordinates (x_i, y_i).

Each move, we can walk one square north, east, south, or west if the square is not in the array of blocked squares. We are also not allowed to walk outside of the grid.

Return true if and only if it is possible to reach the target square from the source square through a sequence of valid moves.

Example 1:

Input: blocked = [[0,1],[1,0]], source = [0,0], target = [0,2]
Output: false
Explanation: The target square is inaccessible starting from the source square because we cannot move.
We cannot move north or east because those squares are blocked.
We cannot move south or west because we cannot go outside of the grid.

Example 2:

Input: blocked = [], source = [0,0], target = [999999,999999]
Output: true
Explanation: Because there are no blocked cells, it is possible to reach the target square.

Constraints:

0 <= blocked.length <= 200
blocked[i].length == 2
0 <= x_i, y_i < 10⁶
source.length == target.length == 2
0 <= s_x, s_y, t_x, t_y < 10⁶
source != target
It is guaranteed that source and target are not blocked.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the dimensions of the maze? Are they bounded by a reasonable maximum?
How is the maze represented? Is it a 2D array of booleans, integers, or some other data structure, and what do the values represent (e.g., 0 for open, 1 for blocked)?
Where are the start and end points located within the maze? Are they guaranteed to be valid and unblocked positions?
If no path exists, what should I return? Should I return null, an empty list/array, or throw an exception?
Are diagonal movements allowed, or can I only move horizontally and vertically?

Brute Force Solution

Approach

The brute force method to escape a maze is like trying every single path until you find one that leads out. We explore all possible directions step-by-step, without being smart about which way to go.

Here's how the algorithm would work step-by-step:

Begin at the starting point within the maze.
Consider all possible directions you can move from your current position (e.g., up, down, left, right).
For each of these directions, imagine taking a step in that direction. If that step leads you to a valid location (not a wall or outside the maze), mark that spot as visited and remember how you got there.
Repeat the process from each of these new locations, again considering all possible directions.
Keep going until you either reach the exit of the maze, or you've visited every possible location in the maze.
If you reach the exit, you have found a path. You can trace back your steps (following the 'how you got there' markers) to find the complete path from the start to the exit.
If you visit every possible location without finding the exit, it means there is no escape.

Code Implementation

def escape_maze_brute_force(maze, start_row, start_column, end_row, end_column):
    rows = len(maze)
    columns = len(maze[0])
    visited = set()
    queue = [(start_row, start_column, [])]

    while queue:
        current_row, current_column, path = queue.pop(0)

        if (current_row, current_column) == (end_row, end_column):
            return path + [(current_row, current_column)]

        if (current_row, current_column) in visited:
            continue

        visited.add((current_row, current_column))

        # Explore possible directions
        possible_moves = [
            (current_row - 1, current_column),  # Up
            (current_row + 1, current_column),  # Down
            (current_row, current_column - 1),  # Left
            (current_row, current_column + 1)   # Right
        ]

        for next_row, next_column in possible_moves:
            # Make sure to stay inside the maze
            if 0 <= next_row < rows and 0 <= next_column < columns and \
               maze[next_row][next_column] == 0:

                # Add valid moves to the queue to explore
                queue.append((next_row, next_column, path + [(current_row, current_column)]))

    return None

Big(O) Analysis

Time Complexity

O(m*n) – The brute force maze escape algorithm explores each possible path in the maze. In the worst-case scenario, we might need to visit every cell in the maze before finding the exit or determining that no exit exists. If the maze has dimensions m rows and n columns, there are m*n cells. Since each cell may be visited, the time complexity is O(m*n), where m and n are the dimensions of the maze.

Space Complexity

O(N) – The algorithm uses a queue (implicitly described by 'remember how you got there' and 'step-by-step') to store the locations to visit, which in the worst case could contain all cells of the maze. Additionally, it needs to keep track of visited locations, potentially using a boolean matrix of the same size as the maze. Therefore, the auxiliary space used depends on the size of the maze, denoted as N. This leads to a space complexity of O(N).

Optimal Solution

Approach

The key to efficiently escaping a large maze is to avoid exploring every possible path. Instead, we'll use a technique that focuses on gradually building a path to the destination, prioritizing expanding outwards from both the start and the target simultaneously to meet in the middle.

Here's how the algorithm would work step-by-step:

Imagine two teams, one starting at the maze's entrance and the other at the exit. Each team is trying to find the other.
Each team takes small steps, exploring the maze outward from their starting point. They mark where they've been to avoid going in circles.
Since the maze is large, we need to avoid getting stuck in dead ends. Limit how far each team explores in each turn to make sure exploration is balanced.
Keep expanding outward from both the start and the end until the two teams find each other. This means they've found a path connecting the entrance and the exit.
If at any point, one of the teams explores a very large area without finding the other or hitting the maze's boundary, it suggests that escaping might be impossible within the given constraints. We will consider this as the maximum area can be explored with available steps.
If the start or target location is enclosed in a small area by obstacles preventing the exploration team to escape to outer boundary, return false. Enclosed area smaller than given steps.
Once the two teams meet, trace the path each team took back to their starting point. This combined path is your escape route.

Code Implementation

def is_escape_possible(blocked, source_point, target_point, max_steps = 20000):    blocked_set = set(map(tuple, blocked))
    # If source or target is blocked, escape is impossible
    if tuple(source_point) in blocked_set or tuple(target_point) in blocked_set:
        return False
    def explore(start_row, start_col, target_row, target_col, blocked_set):        visited = set()
        queue = [(start_row, start_col)]
        visited.add((start_row, start_col))
        # Limit exploration to avoid infinite loops in enclosed areas
        for _ in range(max_steps):
            if not queue:
                return False
            current_row, current_col = queue.pop(0)
            if abs(current_row - target_row) + abs(current_col - target_col) <= 1:
                return True
            for row_direction, col_direction in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                new_row, new_col = current_row + row_direction, current_col + col_direction
                if (new_row, new_col) not in visited and \
                   0 <= new_row <= 10**6 - 1 and 0 <= new_col <= 10**6 - 1 and \
                   (new_row, new_col) not in blocked_set:
                    queue.append((new_row, new_col))
                    visited.add((new_row, new_col))
        return True
    # Explore from source towards target and vice versa
    return explore(source_point[0], source_point[1], target_point[0], target_point[1], blocked_set) and \
           explore(target_point[0], target_point[1], source_point[0], source_point[1], blocked_set)

Big(O) Analysis

Time Complexity

O(p²) – The time complexity is determined by the area explored by the two teams expanding from the start and target locations. Let 'p' represent the maximum number of steps (or the limit on exploration distance) allowed for each team. In the worst-case scenario, each team explores an area proportional to p². Therefore, the total area explored by both teams is 2 * p², which simplifies to O(p²). The algorithm stops when the two explored areas meet or when the exploration limit 'p' is reached.

Space Complexity

O(N) – The algorithm uses two sets to keep track of visited locations from the start and target, respectively. In the worst-case scenario, the exploration teams might visit most of the maze cells before meeting or determining it's impossible to escape. Where N is the total number of cells in the maze, these sets could potentially store close to N cell coordinates. Therefore, the auxiliary space used scales linearly with the size of the maze.

Edge Cases

Case	How to Handle
Maze is null or empty	Return false immediately as there's no maze to escape from, indicating escape impossible.
Start or target is out of bounds	Return false immediately since the starting or target position is invalid.
Start and target are the same cell	Return true immediately because no path is needed; escape is already achieved.
Maze is fully blocked (all cells are walls)	Return false immediately as there is no possible path to escape.
Maze dimensions are very large (potential memory issues)	Consider using an iterative BFS approach with limited queue size or A* to manage memory usage, preventing OOM errors.
Start or target is a wall	Return false immediately because you cannot start or end on a blocked cell.
Integer overflow when calculating distances or coordinates	Use appropriate data types (e.g., long) to store distances or coordinates, or handle calculations modulo a large prime number.
Cycles in the search path (infinite loop)	Maintain a 'visited' set to prevent revisiting the same cell, ensuring termination and correct result.