What is the time complexity of Making A Large Island?

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

What is the space complexity of Making A Large Island?

The space complexity is O(n^2).

What data structures are used to solve Making A Large Island?

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

What are the key insights for solving Making A Large Island?

Insight 1: The core idea is to identify and label each existing island in the grid with a unique ID and store the size of each island. This is done so we can quickly calculate the new island size after changing a 0 to a 1. Insight 2: When encountering a 0, we need to check its four neighbors to see if they belong to different islands. By summing the sizes of these neighboring islands (and adding 1 for the newly converted cell), we can determine the potential size of a new merged island.

Which companies ask LeetCode 827. Making A Large Island in interviews?

This problem is frequently asked at Airbnb, Anduril, DoorDash, Snap, Snowflake and 2 more companies.

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Making A Large Island - LeetCode 827 Solution

Making A Large Island - Complete Solution Guide

Making A Large Island is LeetCode problem 827, 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

You are given an n x n binary matrix grid . You are allowed to change at most one 0 to be 1 . Return the size of the largest island in grid after applying this operation . An island is a 4-directionally connected group of 1 s. Example 1: Input: grid = [[1,0],[0,1]] Output: 3 Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3. Example 2: Input: grid = [[1,1],[1,0]] Output: 4 Explanation: Change the 0 to 1 and make the island bigger, only one island with area =

Detailed Explanation

The problem asks us to find the largest possible island in a given n x n binary matrix (grid) after changing at most one 0 to a 1. An island is defined as a group of 1s that are 4-directionally connected (up, down, left, right). We need to return the size of the largest island after the potential change.

Solution Approach

The provided solution uses a Depth-First Search (DFS) approach to identify and label each island with a unique ID. It then iterates through the grid, and when a 0 is encountered, it checks its neighbors to determine if they belong to different islands. If they do, it calculates the potential size of the new island by summing the sizes of the neighboring islands and adding 1 (for the flipped zero). Finally, it returns the maximum island size found.

Step-by-Step Algorithm

Step 1: Iterate through the grid and identify each island using DFS. During DFS, assign a unique ID to each island and store the size of the island in a hash map/array. This is done by changing each 1 in the grid to its corresponding island ID.
Step 2: Iterate through the grid again. When a 0 is encountered, check its four neighbors (up, down, left, right).
Step 3: For each 0, find the unique IDs of the neighboring islands. Use a set to avoid double-counting islands if a single island borders the 0 in multiple places.
Step 4: Calculate the potential island size by summing the sizes of the neighboring islands (retrieved from the hash map) and adding 1 (for the flipped 0).
Step 5: Update the maximum island size found so far.
Step 6: Return the maximum island size.

Key Insights

Insight 1: The core idea is to identify and label each existing island in the grid with a unique ID and store the size of each island. This is done so we can quickly calculate the new island size after changing a 0 to a 1.
Insight 2: When encountering a 0, we need to check its four neighbors to see if they belong to different islands. By summing the sizes of these neighboring islands (and adding 1 for the newly converted cell), we can determine the potential size of a new merged island.
Insight 3: We must handle the edge case where there are no zeros in the grid. In this case, the largest island is simply the size of the original grid (which is the size of the one island). We should also handle the edge case where there are no ones, which means the largest island size will be 1, because we are allowed to flip at most one zero to one.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

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

Companies

Asked at: Airbnb, Anduril, DoorDash, Snap, Snowflake, UiPath, jio.

Making A Large Island | LeetCode 827

Hard

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

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.Set;
import java.util.Stack;

class Solution {
    public int largestIsland(int[][] grid) {
        int n = grid.length;
        
        class Pair {
            int row;
            int col;

            public Pair(int row, int col) {
                this.row = row;
                this.col = col;
            }
        }

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Problem Info

Description

Problem 827: Making A Large Island - Detailed explanation and solution approaches.

Topics

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

Companies

AirbnbAndurilDoorDashSnapSnowflakeUiPathjio

Stats

Difficulty:Hard

Languages:4

Companies:7

Problem #:827

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

6 steps

1
Step 1: Iterate through the grid and identify each island using DFS. During DFS, assign a unique ID to each island and store the size of the island in a hash map/array. This is done by changing each 1 in the grid to its corresponding island ID.
2
Step 2: Iterate through the grid again. When a 0 is encountered, check its four neighbors (up, down, left, right).
3
Step 3: For each 0, find the unique IDs of the neighboring islands. Use a set to avoid double-counting islands if a single island borders the 0 in multiple places.
4
Step 4: Calculate the potential island size by summing the sizes of the neighboring islands (retrieved from the hash map) and adding 1 (for the flipped 0).
5
Step 5: Update the maximum island size found so far.
6
Step 6: Return the maximum island size.

💡

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 two nested loops that iterate over the entire grid, making it O(n^2). The space complexity is due to the storage of island sizes (potentially O(n^2) unique islands) and the stack used during DFS, which, in the worst case, can also reach O(n^2) when the whole grid is one big island.

📖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

Union Find

Use the Union Find data structure to connect adjacent 1s into islands. Then, iterate through the 0s and calculate the size of the potential island by unioning the connected components of the neighboring 1s.

Time: O(n^2 * α(n^2)), where α is the inverse Ackermann function, which grows very slowly and can be considered almost constant. The Union Find operations are nearly O(1) on average.Space: O(n^2) to store the parent array in the Union Find data structure.

Trade-offs: Union Find might be slightly more complex to implement but can potentially offer better performance in cases where there are many small islands due to the near-constant time complexity of union and find operations.

✨

Key Insights

1Insight 1: The core idea is to identify and label each existing island in the grid with a unique ID and store the size of each island. This is done so we can quickly calculate the new island size after changing a 0 to a 1.
2Insight 2: When encountering a 0, we need to check its four neighbors to see if they belong to different islands. By summing the sizes of these neighboring islands (and adding 1 for the newly converted cell), we can determine the potential size of a new merged island.
3Insight 3: We must handle the edge case where there are no zeros in the grid. In this case, the largest island is simply the size of the original grid (which is the size of the one island). We should also handle the edge case where there are no ones, which means the largest island size will be 1, because we are allowed to flip at most one zero to one.

📚

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 grid contains only 1s. In this case, the algorithm correctly identifies the entire grid as one island and returns its size (n*n).
Edge case 2: The grid contains only 0s. The algorithm will iterate through all zeros and check neighbors. The largest island size will be 1 (flipping one zero to one)
Edge case 3: The grid is very small (e.g., 1x1 or 2x2). The algorithm correctly handles these cases by iterating and performing DFS as needed.
Edge case 4: Two islands are very close, such that flipping a single 0 connects them. The algorithm correctly identifies and merges those islands.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to use a set to store neighboring island IDs when considering a 0. This can lead to double-counting the size of an island if it borders the 0 in multiple places.

✗Common mistake 2: Not handling the case where the grid contains only 1s or only 0s correctly. Failing to do so can lead to incorrect results or runtime errors.

✗Common mistake 3: Incorrectly implementing the DFS, leading to incomplete or incorrect island identification and size calculation.

✅

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 connected regions (islands) of pixels with the same color or intensity.

🏦Real-world scenario 2: Network analysis - Finding connected components in a network graph, which can represent clusters of users or devices that are related.

🎮Real-world scenario 3: Resource allocation - Determining the size and connectivity of available land parcels for development or conservation projects.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ 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: September 16, 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 14

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

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 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.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Making A Large Island"?

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

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