#1036

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Escape a Large Maze

The 'Escape a Large Maze' problem involves navigating a massive grid with blocked squares and finding if a target can be reached from a source.

array hash table depth first search breadth first search

Problem Statement

You are given a grid with coordinates ranging from (0, 0) to (10^6, 10^6). Each square in the grid can either be blocked or open. Your task is to determine if it’s possible to travel from a given source square to a target square, avoiding blocked squares and staying within the grid’s boundaries. The moves allowed are north, south, east, and west, but you cannot move outside the grid or into blocked squares.

The grid is represented by a list of blocked squares, where each element is a pair of coordinates representing a blocked square. The source and target are guaranteed to be open. If there’s a path to the target, return true; otherwise, return false. The problem tests your ability to efficiently navigate large grids while considering blocked paths.

Examples

Example 1

Input: blocked = [[0,1],[1,0]], source = [0,0], target = [0,2]

Output: false

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

Because there are no blocked cells, it is possible to reach the target square.

Constraints

0 <= blocked.length <= 200
blocked[i].length == 2
0 <= xi, yi < 106
source.length == target.length == 2
0 <= sx, sy, tx, ty < 106
source != target
It is guaranteed that source and target are not blocked.

Solution Approach

Array Scanning and Hash Lookup

Begin by scanning the grid using array traversal methods. Maintain a hash set for blocked squares to ensure O(1) lookup time when checking if a square is blocked. This reduces the time complexity and prevents inefficient checks.

Graph Traversal with BFS/DFS

Use Breadth-First Search (BFS) or Depth-First Search (DFS) to explore reachable squares from the source. BFS is preferable for ensuring the shortest path, but DFS can also be used depending on the scenario. Either approach helps avoid revisiting previously explored squares and finding a valid path.

Handling Edge Cases and Loops

Ensure that your algorithm handles situations where a loop is created around the source or the target. If there is no way to progress or if you are stuck in a loop, it’s impossible to reach the target. Efficiently managing explored paths is key to avoiding such loops.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time and space complexity depend on the approach used. BFS typically has O(V + E) complexity where V is the number of squares and E is the number of edges (possible moves). Space complexity is proportional to the number of visited squares, which may also be O(V). With hash set optimizations, the time complexity for blocked square lookups can be reduced to O(1).

What Interviewers Usually Probe

The candidate effectively handles blocked squares with hash lookups.
The candidate avoids revisiting squares and correctly uses BFS/DFS to explore the grid.
The candidate efficiently manages edge cases, such as blocked loops around the source or target.

Common Pitfalls or Variants

Common pitfalls

Failing to account for the case where a path is blocked by surrounding obstacles, causing unnecessary loops.
Inefficient space or time usage, especially when the grid is large, leading to excessive memory consumption or slow performance.
Incorrectly assuming the target is unreachable due to insufficient exploration or misunderstanding of the BFS/DFS traversal method.

Follow-up variants

Implementing an optimized DFS that only explores viable paths to avoid excessive backtracking.
Using BFS with a priority queue to prioritize movement in certain directions.
Solving the problem by implementing a dynamic programming approach that considers each possible move.

FAQ

What is the key pattern for solving the 'Escape a Large Maze' problem?

The main pattern involves using array scanning in combination with hash lookups to efficiently navigate a grid while avoiding blocked squares.

Can BFS or DFS be used to solve the 'Escape a Large Maze' problem?

Yes, both BFS and DFS are suitable for exploring the grid. BFS is generally better for finding the shortest path, while DFS can be useful for specific scenarios.

What happens if I get stuck in a loop while solving this problem?

If you are stuck in a loop, it means that your algorithm is revisiting squares unnecessarily. Proper management of visited squares will help avoid this issue.

How does the problem scale with a 1 million by 1 million grid?

The problem scales by leveraging efficient search techniques such as BFS/DFS and hash lookups. Without proper optimization, the problem becomes infeasible due to the large number of squares.

Are there any variants or optimizations for the 'Escape a Large Maze' problem?

Yes, variants include using dynamic programming or priority queues to improve performance in specific cases. Optimizations focus on reducing redundant calculations during traversal.

terminal

Solution

Solution 1: DFS

The problem can be interpreted as determining whether it is possible to move from a source point to a target point in a $10^6 \times 10^6$ grid, given a small number of blocked points.

class Solution:
    def isEscapePossible(
        self, blocked: List[List[int]], source: List[int], target: List[int]
    ) -> bool:
        def dfs(source: List[int], target: List[int], vis: set) -> bool:
            vis.add(tuple(source))
            if len(vis) > m:
                return True
            for a, b in pairwise(dirs):
                x, y = source[0] + a, source[1] + b
                if 0 <= x < n and 0 <= y < n and (x, y) not in s and (x, y) not in vis:
                    if [x, y] == target or dfs([x, y], target, vis):
                        return True
            return False

        s = {(x, y) for x, y in blocked}
        dirs = (-1, 0, 1, 0, -1)
        n = 10**6
        m = len(blocked) ** 2 // 2
        return dfs(source, target, set()) and dfs(target, source, set())

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