What is the time complexity of Escape a Large Maze?

The optimal solution has a time complexity of O(N^2).

What is the space complexity of Escape a Large Maze?

The space complexity is O(N^2).

What data structures are used to solve Escape a Large Maze?

This problem uses Array, Hash Table, Depth-First Search, Breadth-First Search. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Escape a Large Maze?

Insight 1: The large grid size makes exploring the entire space infeasible. We need to find a way to determine escape without exploring the entire grid. Insight 2: The small number of blocked cells implies that if we explore a region larger than the maximum area that can be enclosed by the blocked cells, we can assume we've 'escaped' to the unbounded open space.

Which companies ask LeetCode 1036. Escape a Large Maze in interviews?

This problem is frequently asked at UiPath.

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Escape a Large Maze - LeetCode 1036 Solution

Q: How to solve Escape a Large Maze?

The solution uses a modified Breadth-First Search (BFS) algorithm. Instead of searching the entire grid, it limits the search to a maximum number of visited cells. The crucial idea is that if the BFS explores more than `N^2` cells (where N is the number of blocked cells), it means the search has escaped the area that could potentially be blocked by the blocked cells and has reached the 'ocean' of open space. The algorithm performs BFS twice: once from the source to the target and another time from the target to the source. The method returns `true` only if both BFS calls can either reach the target/source or escape to the open area.

Escape a Large Maze - Complete Solution Guide

Escape a Large Maze is LeetCode problem 1036, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to determine if it's possible to escape a large maze (1 million x 1 million grid) from a given source coordinate to a target coordinate, given a list of blocked cells. We can move one step in any of the four cardinal directions (north, east, south, west), but we cannot move onto blocked cells or outside the grid. The key challenge is that the grid is very large, making a standard BFS or DFS approach potentially inefficient if it explores the entire grid. Crucially, we are given that the number of blocked cells is relatively small (<= 200).

Solution Approach

The solution uses a modified Breadth-First Search (BFS) algorithm. Instead of searching the entire grid, it limits the search to a maximum number of visited cells. The crucial idea is that if the BFS explores more than `N^2` cells (where N is the number of blocked cells), it means the search has escaped the area that could potentially be blocked by the blocked cells and has reached the 'ocean' of open space. The algorithm performs BFS twice: once from the source to the target and another time from the target to the source. The method returns `true` only if both BFS calls can either reach the target/source or escape to the open area.

Step-by-Step Algorithm

Step 1: Create a set of blocked cells for quick lookup during the BFS.
Step 2: Calculate the maximum number of cells that can be enclosed by blocked cells, which is roughly `N^2 / 2`. The code uses `N^2` as a safe upper bound.
Step 3: Implement the BFS algorithm. It starts from a given source and explores the grid in all four directions.
Step 4: During the BFS, check if the current cell is the target. If so, return `true`.
Step 5: During the BFS, check if the number of visited cells exceeds the maximum allowed (`N^2`). If so, return `true` (escape condition).
Step 6: If the BFS completes without finding the target or escaping (queue becomes empty), it means the source is trapped in a small area. Return `false` in this case.
Step 7: Call the BFS twice: once from the source to the target and once from the target to the source. Return `true` if and only if both calls return `true`.

Key Insights

Insight 1: The large grid size makes exploring the entire space infeasible. We need to find a way to determine escape without exploring the entire grid.
Insight 2: The small number of blocked cells implies that if we explore a region larger than the maximum area that can be enclosed by the blocked cells, we can assume we've 'escaped' to the unbounded open space.
Insight 3: The problem needs two BFS calls - one starting from the source and targeting the target, and another starting from the target and targeting the source. We return `true` only when both searches are successful. This is critical because a one-way path isn't sufficient.
Insight 4: The constraint source != target needs to be considered, though not explicitly coded in the given solutions, it helps in understanding the problem domain.

Complexity Analysis

Time Complexity: O(N^2)

Space Complexity: O(N^2)

Topics

This problem involves: Array, Hash Table, Depth-First Search, Breadth-First Search.

Companies

Asked at: UiPath.

Escape a Large Maze | LeetCode 1036

Hard

Array Hash Table Depth-First Search Breadth-First Search

Time: O(N^2)Space: O(N^2)

Solution Code

Available in:

import java.util.*;

class Solution {
    public boolean isEscapePossible(int[][] blocked, int[] source, int[] target) {
        if (blocked.length == 0) {
            return true;
        }

        Set<String> blockedSet = new HashSet<>();
        for (int[] p : blocked) {
            blockedSet.add(p[0] + "," + p[1]);
        }

        int GRID_SIZE = 1_000_000;
        int MAX_VISITED = blocked.length * blocked.length;

        String sourceTuple = source[0] + "," + source[1];
        String targetTuple = target[0] + "," + target[1];

        
        
        

        // Helper function to convert int[] to string key
        

        
        
        

        

        

        
        
        

        

        

        

        

        

        

        

        

        

        boolean sourceEscapes = canReach(source, target, blockedSet, GRID_SIZE, MAX_VISITED);
        boolean targetEscapes = canReach(target, source, blockedSet, GRID_SIZE, MAX_VISITED);
        
        return sourceEscapes && targetEscapes;

    }

    private boolean canReach(int[] start, int[] end, Set<String> blockedSet, int gridSize, int maxVisited) {
        Queue<int[]> queue = new LinkedList<>();
        queue.offer(start);
        Set<String> visited = new HashSet<>();
        visited.add(start[0] + "," + start[1]);

        while (!queue.isEmpty()) {
            if (visited.size() > maxVisited) {
                return true;
            }

            int[] curr = queue.poll();
            int x = curr[0];
            int y = curr[1];

            if (x == end[0] && y == end[1]) {
                return true;
            }

            int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
            for (int[] dir : directions) {
                int nx = x + dir[0];
                int ny = y + dir[1];

                if (nx >= 0 && nx < gridSize && ny >= 0 && ny < gridSize &&
                    !blockedSet.contains(nx + "," + ny) && !visited.contains(nx + "," + ny)) {
                    visited.add(nx + "," + ny);
                    queue.offer(new int[]{nx, ny});
                }
            }
        }
        return false;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1036: Escape a Large Maze - Detailed explanation and solution approaches.

Topics

Array Hash Table Depth-First Search Breadth-First Search

Companies

UiPath

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1036

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Create a set of blocked cells for quick lookup during the BFS.
2
Step 2: Calculate the maximum number of cells that can be enclosed by blocked cells, which is roughly `N^2 / 2`. The code uses `N^2` as a safe upper bound.
3
Step 3: Implement the BFS algorithm. It starts from a given source and explores the grid in all four directions.
4
Step 4: During the BFS, check if the current cell is the target. If so, return `true`.
5
Step 5: During the BFS, check if the number of visited cells exceeds the maximum allowed (`N^2`). If so, return `true` (escape condition).
6
Step 6: If the BFS completes without finding the target or escaping (queue becomes empty), it means the source is trapped in a small area. Return `false` in this case.
7
Step 7: Call the BFS twice: once from the source to the target and once from the target to the source. Return `true` if and only if both calls return `true`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N^2)

💾

Space Complexity

O(N^2)

📝

Why?

The time complexity is determined by the BFS, which explores at most N^2 cells in both directions (from source to target and target to source). The space complexity comes from storing the visited cells and the cells in the queue during BFS, both of which can grow to O(N^2). The blocked set requires O(N) space. The two BFS calls are performed sequentially, not concurrently.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

A* Search

A* search could be used, potentially with the Manhattan distance as the heuristic. However, it doesn't fundamentally change the approach needed to handle the large grid and the need to detect when an 'escape' to unbounded space has been made. It can be more efficient if the heuristic is very effective in directing the search towards the target.

Time: O(b^d) in the worst case, where b is the branching factor and d is the depth of the solution, but can be much better with a good heuristic.Space: O(b^d) in the worst case.

Trade-offs: A* could be better if a good heuristic can be found to guide the search more directly to the target. Without a good heuristic, the performance might be similar or worse than BFS, and it's more complex to implement correctly.

✨

Key Insights

1Insight 1: The large grid size makes exploring the entire space infeasible. We need to find a way to determine escape without exploring the entire grid.
2Insight 2: The small number of blocked cells implies that if we explore a region larger than the maximum area that can be enclosed by the blocked cells, we can assume we've 'escaped' to the unbounded open space.
3Insight 3: The problem needs two BFS calls - one starting from the source and targeting the target, and another starting from the target and targeting the source. We return `true` only when both searches are successful. This is critical because a one-way path isn't sufficient.
4Insight 4: The constraint source != target needs to be considered, though not explicitly coded in the given solutions, it helps in understanding the problem domain.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: No blocked cells. The solution correctly handles this case by returning `true` immediately (as it's always possible to reach the target if there are no obstacles).
Edge case 2: Source and target are adjacent, but blocked cells surround them. The BFS will explore these adjacent cells and may either find a path or hit the `MAX_VISITED` limit. It will correctly determine if escape is possible within the limited search space.
Edge case 3: Source and target are separated by a large area but are still connectable by some winding path in the unblocked area, the BFS will explore the space until it connects or hits `MAX_VISITED` limit.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking for escape from both source and target. Failing to check both directions will incorrectly identify paths when only one direction is possible.

✗Common mistake 2: Not handling the case where there are no blocked cells properly. This can cause incorrect results or runtime errors.

✗Common mistake 3: Forgetting to limit the BFS exploration. Exploring the entire grid is not feasible and will lead to a timeout.

✗Common mistake 4: Using the wrong data structures for blocked and visited which can impact the runtime significantly. HashSets are necessary for constant time lookup.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Pathfinding in large open-world games with sparsely populated obstacles. This approach could be applied when finding paths between two points with a limited number of static obstacles on a very large map. The 'escape' condition helps to avoid exploring vast open spaces unnecessarily.

🏦Real-world scenario 2: Network routing with blacklisted nodes. Imagine finding a route between two servers in a large network, where a small number of routers are temporarily unavailable. The blacklisted routers are analogous to the 'blocked' cells, and the same approach can be used to efficiently find a path if one exists.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 15, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Escape a Large Maze"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?