What is the time complexity of Max Area of Island?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Max Area of Island?

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

What are the key insights for solving Max Area of Island?

Insight 1: The problem can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS) to explore each island. Insight 2: To avoid infinite loops and double-counting, we need to mark visited cells (cells with value 1) as visited by changing their value to 0 after encountering them.

Which companies ask LeetCode 695. Max Area of Island in interviews?

This problem is frequently asked at Disney, DoorDash, Dropbox, Grubhub, Intuit and 6 more companies.

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

Max Area of Island - LeetCode 695 Solution

Q: What is the space complexity of Max Area of Island?

The space complexity is O(m*n).

Max Area of Island - Complete Solution Guide

Max Area of Island is LeetCode problem 695, 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

You are given an m x n binary matrix grid . An island is a group of 1 's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water. The area of an island is the number of cells with a value 1 in the island. Return the maximum area of an island in grid . If there is no island, return 0 . Example 1: Input: grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1

Detailed Explanation

The problem asks us to find the maximum area of an island in a given binary matrix (grid). An island is defined as a group of '1's that are connected 4-directionally (up, down, left, right). The input is a 2D array (grid) of 0s and 1s, where 1 represents land and 0 represents water. The output should be the largest number of connected 1s in the grid. If there are no islands (no 1s), the output should be 0.

Solution Approach

The solution uses Depth-First Search (DFS) to explore each island and calculate its area. The algorithm iterates through each cell of the grid. If a cell contains a '1', it means a new island is encountered. A DFS is initiated from that cell to explore the entire island. During the DFS, each visited '1' is changed to '0' to mark it as visited and prevent revisiting. The area of the island is calculated during the traversal. The maximum area found so far is updated.

Step-by-Step Algorithm

Step 1: Initialize a variable `max_area` to 0 to store the maximum island area found so far.
Step 2: Iterate through each cell in the `grid` using nested loops.
Step 3: If the current cell `grid[i][j]` is equal to 1, then it represents a starting point of an island.
Step 4: Initialize `current_area` to 0 and create a stack to perform DFS. Push the starting cell (i, j) onto the stack.
Step 5: Change the value of the current cell `grid[i][j]` to 0 to mark it as visited. This is crucial to prevent infinite loops.
Step 6: While the stack is not empty:
a. Pop a cell (r, c) from the stack.
b. Increment `current_area` by 1.
c. Explore the 4-directional neighbors (up, down, left, right) of the current cell (r, c).
d. For each neighbor (nr, nc), check if it is within the grid boundaries and if its value is 1 (unvisited land).
e. If the neighbor is a valid unvisited land, push it onto the stack and mark it as visited by setting `grid[nr][nc]` to 0.
Step 7: After the DFS is complete for the current island, update `max_area` to be the maximum of `max_area` and `current_area`.
Step 8: After iterating through all the cells in the `grid`, return `max_area`.

Key Insights

Insight 1: The problem can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS) to explore each island.
Insight 2: To avoid infinite loops and double-counting, we need to mark visited cells (cells with value 1) as visited by changing their value to 0 after encountering them.
Insight 3: Iterate through the grid, and whenever you find a '1', start a DFS/BFS to explore the island connected to that cell. Keep track of the largest area found so far.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

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

Companies

Asked at: Disney, DoorDash, Dropbox, Grubhub, Intuit, LinkedIn, Salesforce, Smartsheet, Snowflake, Tesla, Wise.

Max Area of Island | LeetCode 695

Medium

Array Depth-First Search Breadth-First Search Union Find Matrix

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        int max_area = 0;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    int current_area = 0;
                    Stack<int[]> stack = new Stack<>();
                    stack.push(new int[]{i, j});
                    grid[i][j] = 0;

                    while (!stack.isEmpty()) {
                        int[] current = stack.pop();
                        int r = current[0];
                        int c = current[1];
                        current_area += 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 < m && 0 <= nc && nc < n && grid[nr][nc] == 1) {
                                stack.push(new int[]{nr, nc});
                                grid[nr][nc] = 0;
                            }
                        }
                    }

                    max_area = Math.max(max_area, current_area);
                }
            }
        }

        return max_area;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 695: Max Area of Island - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search Union Find Matrix

Companies

DisneyDoorDashDropboxGrubhubIntuitLinkedInSalesforceSmartsheet

Stats

Difficulty:Medium

Languages:4

Companies:11

Problem #:695

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

13 steps

1
Step 1: Initialize a variable `max_area` to 0 to store the maximum island area found so far.
2
Step 2: Iterate through each cell in the `grid` using nested loops.
3
Step 3: If the current cell `grid[i][j]` is equal to 1, then it represents a starting point of an island.
4
Step 4: Initialize `current_area` to 0 and create a stack to perform DFS. Push the starting cell (i, j) onto the stack.
5
Step 5: Change the value of the current cell `grid[i][j]` to 0 to mark it as visited. This is crucial to prevent infinite loops.
6
Step 6: While the stack is not empty:
7
a. Pop a cell (r, c) from the stack.
8
b. Increment `current_area` by 1.
9
c. Explore the 4-directional neighbors (up, down, left, right) of the current cell (r, c).
10
d. For each neighbor (nr, nc), check if it is within the grid boundaries and if its value is 1 (unvisited land).
11
e. If the neighbor is a valid unvisited land, push it onto the stack and mark it as visited by setting `grid[nr][nc]` to 0.
12
Step 7: After the DFS is complete for the current island, update `max_area` to be the maximum of `max_area` and `current_area`.
13
Step 8: After iterating through all the cells in the `grid`, return `max_area`.

💡

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(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is O(m*n) because we iterate through each cell in the grid. The space complexity is O(m*n) in the worst-case scenario when the whole grid is filled with 1s which will then require a stack of size O(m*n).

📖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

Breadth-First Search (BFS)

Instead of using DFS with a stack, we can use BFS with a queue. The core logic remains the same: explore the island, mark visited cells, and calculate the area.

Time: O(m*n)Space: O(m*n)

Trade-offs: The time and space complexity are the same as DFS. The choice between BFS and DFS is often a matter of preference or coding style. BFS may be preferable in some scenarios due to its iterative nature, potentially avoiding stack overflow issues for extremely large grids, though the space complexity is the same.

Union Find

Treat each cell as a node and connect adjacent land cells (1s) with a Union Find data structure. Each connected component represents an island. The size of each component can be tracked, and the maximum size is the answer.

Time: O(m*n * α(m*n)), where α is the inverse Ackermann function, which grows very slowly, so it's practically constant.Space: O(m*n)

Trade-offs: Union Find can be more complex to implement than DFS or BFS for this specific problem. While the time complexity is slightly better in theory, the difference is negligible in practice given the problem constraints. DFS and BFS are generally easier to understand and implement for this problem.

✨

Key Insights

1Insight 1: The problem can be solved using Depth-First Search (DFS) or Breadth-First Search (BFS) to explore each island.
2Insight 2: To avoid infinite loops and double-counting, we need to mark visited cells (cells with value 1) as visited by changing their value to 0 after encountering them.
3Insight 3: Iterate through the grid, and whenever you find a '1', start a DFS/BFS to explore the island connected to that cell. Keep track of the largest area found so far.

📚

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: Empty grid (grid with no rows or columns). The code implicitly handles this as the loops won't execute, and the initial `max_area` of 0 will be returned.
Edge case 2: Grid with only 0s (no islands). The code correctly returns 0 because the DFS is never initiated.
Edge case 3: Grid with a single island. The code correctly identifies and calculates the area of the single island.
Edge case 4: Grid with multiple disconnected islands of varying sizes. The code correctly finds and calculates the area of each island and returns the maximum area.
Edge case 5: A very large grid. The code handles this efficiently within the problem constraints due to its O(m*n) complexity.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not marking visited cells as visited. This leads to infinite loops as the DFS/BFS will keep revisiting the same cells.

✗Common mistake 2: Not checking the boundaries of the grid when exploring neighbors. This can lead to array out-of-bounds errors.

✗Common mistake 3: Incorrectly calculating the area of the island by not incrementing the area counter in each step of the DFS/BFS.

✗Common mistake 4: Forgetting to update the max_area after each island's area is calculated.

✗Common mistake 5: Confusing row and column indices when accessing grid elements.

✅

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: Image processing: Identifying and measuring the size of connected regions of pixels in an image. For example, identifying objects in satellite imagery or medical scans.

🏦Real-world scenario 2: Network analysis: Finding the largest connected component in a network graph. This can be used to analyze social networks or infrastructure networks.

🎮Real-world scenario 3: Game development: Determining the size of territories in a strategy game.

📱Real-world scenario 4: Pathfinding: Identifying connected areas for determining traversable regions in a map.

💡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.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: November 3, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

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

@AlgoMasterJul 7

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

@InterviewProJun 30

❓

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 "Max Area of Island"?

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 depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?