What is the time complexity of As Far from Land as Possible?

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

What is the space complexity of As Far from Land as Possible?

The space complexity is O(n^2).

What data structures are used to solve As Far from Land as Possible?

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

What are the key insights for solving As Far from Land as Possible?

Insight 1: Using Breadth-First Search (BFS) is efficient for finding the shortest distance from multiple source points (land cells) to all water cells. Insight 2: Treating all land cells as the starting points for a multi-source BFS helps find the maximum distance to the farthest water cell.

Which companies ask LeetCode 1162. As Far from Land as Possible in interviews?

This problem is frequently asked at UiPath, Wix.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

As Far from Land as Possible - LeetCode 1162 Solution

As Far from Land as Possible - Complete Solution Guide

As Far from Land as Possible is LeetCode problem 1162, 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 grid containing only values 0 and 1 , where 0 represents water and 1 represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance. If no land or water exists in the grid, return -1 . The distance used in this problem is the Manhattan distance: the distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1| . Example 1: Input: grid = [[1,0,1],[0,0,0],[1,0,1]] Output: 2 Explanation: The cell (1, 1) is as far a

Detailed Explanation

The problem asks us to find the maximum Manhattan distance from a water cell (value 0) to the nearest land cell (value 1) in a given n x n grid. The Manhattan distance between two cells (x0, y0) and (x1, y1) is |x0 - x1| + |y0 - y1|. If the grid contains no land or no water, we should return -1.

Solution Approach

The solution uses a multi-source Breadth-First Search (BFS) to find the maximum distance from a water cell to the nearest land cell. First, we enqueue all land cells into the queue. Then, we perform a standard BFS. Each layer of the BFS represents an increment in distance. We explore the grid layer by layer, marking visited water cells and adding them to the queue for the next level. We keep track of the distance and update it after each layer. The algorithm stops when the queue is empty, which means all reachable water cells have been visited. The final distance is the maximum distance from any water cell to its nearest land cell.

Step-by-Step Algorithm

Step 1: Initialize a queue and enqueue all land cells (cells with value 1) from the grid.
Step 2: Handle edge cases: If the queue is empty (no land) or the queue contains all the cells in the grid (no water), return -1.
Step 3: Initialize the distance to -1. This will be incremented at the start of each BFS level.
Step 4: Perform a BFS: While the queue is not empty:
Step 5: Increment the distance by 1 (start of a new layer/distance).
Step 6: Get the size of the queue (number of cells to process at the current distance).
Step 7: Iterate 'q_size' times. In each iteration:
Step 8: Dequeue a cell (r, c) from the queue.
Step 9: Explore the four adjacent neighbors (up, down, left, right).
Step 10: For each neighbor (nr, nc):
Step 11: If the neighbor is within the grid boundaries and is a water cell (value 0):
Step 12: Mark the neighbor as visited by changing its value in the grid to 1.
Step 13: Enqueue the neighbor into the queue.
Step 14: After the BFS is complete, return the final distance.

Key Insights

Insight 1: Using Breadth-First Search (BFS) is efficient for finding the shortest distance from multiple source points (land cells) to all water cells.
Insight 2: Treating all land cells as the starting points for a multi-source BFS helps find the maximum distance to the farthest water cell.
Insight 3: The problem can be visualized as expanding outwards from the land cells in all four directions, layer by layer, until all reachable water cells are visited.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

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

Companies

Asked at: UiPath, Wix.

As Far from Land as Possible | LeetCode 1162

Medium

Array Dynamic Programming 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 maxDistance(int[][] grid) {
        int n = grid.length;
        Queue<int[]> q = new LinkedList<>();

        for (int r = 0; r < n; r++) {
            for (int c = 0; c < n; c++) {
                if (grid[r][c] == 1) {
                    q.offer(new int[]{r, c});
                }
            }
        }

        if (q.isEmpty() || q.size() == n * n) {
            return -1;
        }

        int distance = -1;
        while (!q.isEmpty()) {
            distance++;
            int q_size = q.size();
            for (int i = 0; i < q_size; i++) {
                int[] curr = q.poll();
                int r = curr[0];
                int c = curr[1];

                int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
                for (int[] dir : directions) {
                    int dr = dir[0];
                    int dc = dir[1];
                    int nr = r + dr;
                    int nc = c + dc;

                    if (0 <= nr && nr < n && 0 <= nc && nc < n && grid[nr][nc] == 0) {
                        grid[nr][nc] = 1;
                        q.offer(new int[]{nr, nc});
                    }
                }
            }
        }

        return distance;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1162: As Far from Land as Possible - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Breadth-First Search Matrix

Companies

UiPathWix

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1162

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

14 steps

1
Step 1: Initialize a queue and enqueue all land cells (cells with value 1) from the grid.
2
Step 2: Handle edge cases: If the queue is empty (no land) or the queue contains all the cells in the grid (no water), return -1.
3
Step 3: Initialize the distance to -1. This will be incremented at the start of each BFS level.
4
Step 4: Perform a BFS: While the queue is not empty:
5
Step 5: Increment the distance by 1 (start of a new layer/distance).
6
Step 6: Get the size of the queue (number of cells to process at the current distance).
7
Step 7: Iterate 'q_size' times. In each iteration:
8
Step 8: Dequeue a cell (r, c) from the queue.
9
Step 9: Explore the four adjacent neighbors (up, down, left, right).
10
Step 10: For each neighbor (nr, nc):
11
Step 11: If the neighbor is within the grid boundaries and is a water cell (value 0):
12
Step 12: Mark the neighbor as visited by changing its value in the grid to 1.
13
Step 13: Enqueue the neighbor into the queue.
14
Step 14: After the BFS is complete, return the final distance.

💡

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 BFS algorithm visits each cell in the grid at most once. The queue can potentially hold all the water cells. Therefore, the time complexity is proportional to the number of cells in the grid, which is n^2. Similarly, the queue in the worst case will hold all the water cells, leading to a space complexity of O(n^2).

📖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

Dynamic Programming

Compute the distance to the nearest land cell for each cell using dynamic programming. Two passes (forward and backward) are needed to correctly account for cells in all directions.

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

Trade-offs: DP offers a non-queue based approach. The implementation may be slightly more complex than BFS. BFS is generally preferred as it's conceptually simpler for this problem.

✨

Key Insights

1Insight 1: Using Breadth-First Search (BFS) is efficient for finding the shortest distance from multiple source points (land cells) to all water cells.
2Insight 2: Treating all land cells as the starting points for a multi-source BFS helps find the maximum distance to the farthest water cell.
3Insight 3: The problem can be visualized as expanding outwards from the land cells in all four directions, layer by layer, until all reachable water cells are visited.

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: Grid contains no land (all water). The algorithm correctly returns -1 because the initial queue is empty.
Edge case 2: Grid contains no water (all land). The algorithm correctly returns -1 because after initializing the queue with all the land, it detects that the queue contains all the cells of the grid. The BFS loop is never entered.
Edge case 3: Single land cell surrounded by water. The algorithm correctly calculates the distances to all reachable water cells.
Edge case 4: Single water cell surrounded by land. The algorithm finds the distance of 1 to the nearest land.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the edge cases where the grid contains no land or no water, resulting in incorrect output or runtime errors.

✗Common mistake 2: Incorrectly calculating the Manhattan distance or mishandling boundary conditions during the BFS exploration.

✗Common mistake 3: Not marking visited water cells during the BFS, leading to infinite loops and incorrect distance calculations. The grid is modified in place to mark visited cells, a crucial optimization.

✗Common mistake 4: Not using a queue for BFS and trying to implement BFS using recursion, leading to stack overflow errors for large grids.

✅

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: Resource allocation: Finding the optimal location for a new facility (e.g., hospital, fire station) to minimize the maximum distance to any residential area.

🏦Real-world scenario 2: Network routing: Determining the maximum delay in a network where data packets are routed from multiple source nodes to destination nodes.

🎮Real-world scenario 3: Image processing: Distance transforms are used in image analysis and computer vision applications, such as finding the minimum distance from each pixel to the nearest object edge.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: August 22, 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

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

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

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.

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "As Far from Land as Possible"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?