What is the time complexity of Shortest Path in Binary Matrix?

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

What is the space complexity of Shortest Path in Binary Matrix?

The space complexity is O(n^2).

What data structures are used to solve Shortest Path in Binary Matrix?

This problem uses Array, Breadth-First Search, Matrix. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Shortest Path in Binary Matrix?

Insight 1: Breadth-First Search (BFS) is the most appropriate algorithm because it guarantees finding the shortest path in an unweighted graph (where each step has the same cost). Insight 2: We need to consider all 8 possible directions (up, down, left, right, and the four diagonals) when searching for adjacent cells.

Which companies ask LeetCode 1091. Shortest Path in Binary Matrix in interviews?

This problem is frequently asked at Airbnb, Intuit, Palo Alto Networks.

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Shortest Path in Binary Matrix - LeetCode 1091 Solution

Shortest Path in Binary Matrix - Complete Solution Guide

Shortest Path in Binary Matrix is LeetCode problem 1091, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an n x n binary matrix grid , return the length of the shortest clear path in the matrix . If there is no clear path, return -1 . A clear path in a binary matrix is a path from the top-left cell (i.e., (0, 0) ) to the bottom-right cell (i.e., (n - 1, n - 1) ) such that: All the visited cells of the path are 0 . All the adjacent cells of the path are 8-directionally connected (i.e., they are different and they share an edge or a corner). The length of a clear path is the number of visited c

Detailed Explanation

The problem asks us to find the shortest path from the top-left corner (0, 0) to the bottom-right corner (n-1, n-1) of an n x n binary matrix. A 'clear path' consists of cells with the value 0, and adjacent cells in the path are 8-directionally connected (meaning they share an edge or a corner). The length of the path is the number of cells visited. If no such path exists, we return -1. The input is the binary matrix `grid`, and the output is the length of the shortest clear path.

Solution Approach

The solution uses Breadth-First Search (BFS) to find the shortest path. We start at the top-left cell and explore its neighbors in all 8 directions. We use a queue to store the cells to visit. Each cell in the queue is associated with its row, column, and the length of the path from the starting cell. As we visit cells, we mark them as visited by changing their value in the grid to 1. We continue until we reach the bottom-right cell or the queue is empty. If we reach the bottom-right cell, we return the path length. If the queue becomes empty and we haven't reached the destination, it means there is no clear path, so we return -1.

Step-by-Step Algorithm

Step 1: Check if the starting or ending cell is blocked (value 1). If so, return -1 because no path is possible.
Step 2: Initialize a queue with the starting cell (0, 0) and path length 1.
Step 3: Mark the starting cell as visited by changing its value in the grid to 1.
Step 4: Define an array of 8 possible directions (row and column offsets).
Step 5: While the queue is not empty:
Step 6: Dequeue a cell from the queue and extract its row, column, and path length.
Step 7: If the dequeued cell is the destination (bottom-right corner), return the path length.
Step 8: Iterate through the 8 directions:
Step 9: Calculate the coordinates of the neighboring cell by adding the direction offsets to the current cell's coordinates.
Step 10: Check if the neighboring cell is within the grid boundaries and is not blocked (value 0) and not visited (value 0 initially, or it's 1 if already visited during search).
Step 11: If the neighboring cell is valid, mark it as visited (set its value in the grid to 1) and enqueue it with an incremented path length.
Step 12: If the queue becomes empty before reaching the destination, return -1 (no path found).

Key Insights

Insight 1: Breadth-First Search (BFS) is the most appropriate algorithm because it guarantees finding the shortest path in an unweighted graph (where each step has the same cost).
Insight 2: We need to consider all 8 possible directions (up, down, left, right, and the four diagonals) when searching for adjacent cells.
Insight 3: Marking visited cells is crucial to avoid infinite loops and redundant calculations. We can modify the input grid itself to keep track of visited cells.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: Array, Breadth-First Search, Matrix.

Companies

Asked at: Airbnb, Intuit, Palo Alto Networks.

Shortest Path in Binary Matrix | LeetCode 1091

Medium

Array Breadth-First Search Matrix

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

Solution Code

Available in:

import java.util.LinkedList;
import java.util.Queue;

class Solution {
    public int shortestPathBinaryMatrix(int[][] grid) {
        int n = grid.length;
        if (grid[0][0] == 1 || grid[n - 1][n - 1] == 1) {
            return -1;
        }

        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{0, 0, 1}); // (row, col, length)
        grid[0][0] = 1; // Mark as visited

        int[][] directions = {
            {-1, -1}, {-1, 0}, {-1, 1},
            {0, -1},          {0, 1},
            {1, -1}, {1, 0}, {1, 1}
        };

        while (!queue.isEmpty()) {
            int[] current = queue.poll();
            int row = current[0];
            int col = current[1];
            int length = current[2];

            if (row == n - 1 && col == n - 1) {
                return length;
            }

            for (int[] direction : directions) {
                int dr = direction[0];
                int dc = direction[1];
                int new_row = row + dr;
                int new_col = col + dc;

                if (0 <= new_row && new_row < n && 0 <= new_col && new_col < n && grid[new_row][new_col] == 0) {
                    grid[new_row][new_col] = 1; // Mark as visited
                    queue.offer(new int[]{new_row, new_col, length + 1});
                }
            }
        }

        return -1;
    }
}

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Problem Info

Description

Problem 1091: Shortest Path in Binary Matrix - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Matrix

Companies

AirbnbIntuitPalo Alto Networks

Stats

Difficulty:Medium

Languages:4

Companies:3

Problem #:1091

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

12 steps

1
Step 1: Check if the starting or ending cell is blocked (value 1). If so, return -1 because no path is possible.
2
Step 2: Initialize a queue with the starting cell (0, 0) and path length 1.
3
Step 3: Mark the starting cell as visited by changing its value in the grid to 1.
4
Step 4: Define an array of 8 possible directions (row and column offsets).
5
Step 5: While the queue is not empty:
6
Step 6: Dequeue a cell from the queue and extract its row, column, and path length.
7
Step 7: If the dequeued cell is the destination (bottom-right corner), return the path length.
8
Step 8: Iterate through the 8 directions:
9
Step 9: Calculate the coordinates of the neighboring cell by adding the direction offsets to the current cell's coordinates.
10
Step 10: Check if the neighboring cell is within the grid boundaries and is not blocked (value 0) and not visited (value 0 initially, or it's 1 if already visited during search).
11
Step 11: If the neighboring cell is valid, mark it as visited (set its value in the grid to 1) and enqueue it with an incremented path length.
12
Step 12: If the queue becomes empty before reaching the destination, return -1 (no path found).

💡

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)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity is determined by the BFS traversal. Each cell is enqueued and dequeued at most once. The space complexity is determined by the maximum size of the queue, which can be proportional to the number of cells in the grid in the worst case.

📖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

Depth-First Search (DFS)

DFS can also be used to find a path, but it does not guarantee the shortest path. It explores one branch as far as possible before backtracking.

Time: O(n^2) in the worst case, but can be exponential if the path is long.Space: O(n^2) in the worst case due to recursion depth.

Trade-offs: DFS is generally less efficient than BFS for finding the shortest path. It may be faster in some specific cases, but its performance is less predictable. It is generally not recommended for this problem.

A* Search

A* search is an informed search algorithm that uses a heuristic to guide the search towards the goal. It is often used when the cost of each step is not uniform or when we have some information about the distance to the goal.

Time: Varies depending on the heuristic. Can be faster than BFS in some cases if a good heuristic is used.Space: Can be higher than BFS due to storing the open and closed sets.

Trade-offs: A* can be more efficient than BFS if a good heuristic is available. However, choosing a suitable heuristic can be challenging, and a poor heuristic can lead to worse performance than BFS. It is more complex to implement than BFS.

✨

Key Insights

1Insight 1: Breadth-First Search (BFS) is the most appropriate algorithm because it guarantees finding the shortest path in an unweighted graph (where each step has the same cost).
2Insight 2: We need to consider all 8 possible directions (up, down, left, right, and the four diagonals) when searching for adjacent cells.
3Insight 3: Marking visited cells is crucial to avoid infinite loops and redundant calculations. We can modify the input grid itself to keep track of visited cells.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The starting cell (0, 0) or ending cell (n-1, n-1) is blocked (value 1). The solution correctly handles this by immediately returning -1.
Edge case 2: There is no clear path from the start to the end. The solution correctly handles this by returning -1 after the BFS algorithm has explored all possible paths without finding the destination.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not checking if the starting or ending cell is blocked before starting the search. This leads to incorrect results.

✗Common mistake 2: Not marking visited cells, which can lead to infinite loops and stack overflow errors.

✗Common mistake 3: Incorrectly implementing the 8-directional movement. Ensuring all 8 possible movements are handled correctly is crucial.

✅

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 games, where characters need to find the shortest path from one location to another in a grid-based environment.

🏦Real-world scenario 2: Robotics, where a robot needs to navigate through a maze or obstacle course. The binary matrix can represent the environment, with 0s representing free space and 1s representing obstacles.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: June 12, 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 13

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

@InterviewProJul 6

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Shortest Path in Binary Matrix"?

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

🤖 Need help with this problem?