What is the time complexity of Find All Groups of Farmland?

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

What is the space complexity of Find All Groups of Farmland?

The space complexity is O(m*n).

What data structures are used to solve Find All Groups of Farmland?

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

What are the key insights for solving Find All Groups of Farmland?

Insight 1: Since farmland groups are rectangular and not adjacent, we can iterate through the matrix and, upon finding a '1', determine the extent of the rectangle it belongs to without worrying about merging adjacent groups. Insight 2: Modifying the original input matrix 'land' by setting visited farmland to '0' avoids double-counting farmland and simplifies the process of finding distinct groups.

Which companies ask LeetCode 1992. Find All Groups of Farmland in interviews?

This problem is frequently asked at Citrix.

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Find All Groups of Farmland - LeetCode 1992 Solution

Find All Groups of Farmland - Complete Solution Guide

Find All Groups of Farmland is LeetCode problem 1992, 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 a 0-indexed m x n binary matrix land where a 0 represents a hectare of forested land and a 1 represents a hectare of farmland. To keep the land organized, there are designated rectangular areas of hectares that consist entirely of farmland. These rectangular areas are called groups . No two groups are adjacent, meaning farmland in one group is not four-directionally adjacent to another farmland in a different group. land can be represented by a coordinate system where the top left

Detailed Explanation

The problem requires identifying rectangular groups of farmland (represented by '1's) within a given binary matrix (named 'land'). The goal is to find the top-left and bottom-right coordinates of each farmland group. The input is a 2D array of 0s and 1s, and the output should be a 2D array where each inner array contains the coordinates [row1, col1, row2, col2] of a farmland group. Crucially, groups are guaranteed to be rectangular and not adjacent to each other. If no farmland exists, return an empty array.

Solution Approach

The provided solution uses a straightforward iterative approach. It scans the input matrix 'land' row by row, column by column. When a cell with value '1' (farmland) is encountered, it marks the top-left corner of a new farmland group. It then extends the rectangle downwards and rightwards to find the bottom-right corner. After identifying a farmland group, it updates the 'land' matrix by setting the cells within the identified rectangle to '0' to avoid re-processing them. Finally, it stores the top-left and bottom-right coordinates of each found farmland group in a result array.

Step-by-Step Algorithm

Step 1: Initialize an empty list (result) to store the coordinates of the farmland groups.
Step 2: Iterate through the 'land' matrix using nested loops (row r1, column c1).
Step 3: For each cell land[r1][c1], check if its value is '1'.
Step 4: If land[r1][c1] is '1', initialize the bottom-right coordinates (r2, c2) to (r1, c1).
Step 5: Extend the rectangle downwards by incrementing r2 as long as r2 + 1 is within bounds and land[r2+1][c1] is '1'.
Step 6: Extend the rectangle rightwards by incrementing c2 as long as c2 + 1 is within bounds and land[r1][c2+1] is '1'.
Step 7: Add the coordinates [r1, c1, r2, c2] to the result list.
Step 8: Mark the identified farmland region as visited by setting all cells within the rectangle (from (r1,c1) to (r2,c2)) to '0'.
Step 9: Continue iterating through the matrix until all cells have been processed.
Step 10: Return the result list containing the coordinates of all identified farmland groups.

Key Insights

Insight 1: Since farmland groups are rectangular and not adjacent, we can iterate through the matrix and, upon finding a '1', determine the extent of the rectangle it belongs to without worrying about merging adjacent groups.
Insight 2: Modifying the original input matrix 'land' by setting visited farmland to '0' avoids double-counting farmland and simplifies the process of finding distinct groups.
Insight 3: The problem inherently involves nested loops to traverse the matrix, and the rectangle detection adds further complexity. Efficiently determining the boundaries of the rectangle is important for performance.

Complexity Analysis

Time Complexity: O(m*n*(m+n))

Space Complexity: O(m*n)

Topics

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

Companies

Asked at: Citrix.

Find All Groups of Farmland | LeetCode 1992

Medium

Array Depth-First Search Breadth-First Search Matrix

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int[][] findFarmland(int[][] land) {
        int m = land.length;
        int n = land[0].length;
        List<int[]> resultList = new ArrayList<>();

        for (int r1 = 0; r1 < m; r1++) {
            for (int c1 = 0; c1 < n; c1++) {
                if (land[r1][c1] == 1) {
                    int r2 = r1;
                    int c2 = c1;

                    while (r2 + 1 < m && land[r2 + 1][c1] == 1) {
                        r2++;
                    }

                    while (c2 + 1 < n && land[r1][c2 + 1] == 1) {
                        c2++;
                    }

                    resultList.add(new int[]{r1, c1, r2, c2});

                    for (int i = r1; i <= r2; i++) {
                        for (int j = c1; j <= c2; j++) {
                            land[i][j] = 0;
                        }
                    }
                }
            }
        }

        int[][] result = new int[resultList.size()][4];
        for (int i = 0; i < resultList.size(); i++) {
            result[i] = resultList.get(i);
        }

        return result;
    }
}

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Problem Info

Description

Problem 1992: Find All Groups of Farmland - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search Matrix

Companies

Citrix

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1992

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

10 steps

1
Step 1: Initialize an empty list (result) to store the coordinates of the farmland groups.
2
Step 2: Iterate through the 'land' matrix using nested loops (row r1, column c1).
3
Step 3: For each cell land[r1][c1], check if its value is '1'.
4
Step 4: If land[r1][c1] is '1', initialize the bottom-right coordinates (r2, c2) to (r1, c1).
5
Step 5: Extend the rectangle downwards by incrementing r2 as long as r2 + 1 is within bounds and land[r2+1][c1] is '1'.
6
Step 6: Extend the rectangle rightwards by incrementing c2 as long as c2 + 1 is within bounds and land[r1][c2+1] is '1'.
7
Step 7: Add the coordinates [r1, c1, r2, c2] to the result list.
8
Step 8: Mark the identified farmland region as visited by setting all cells within the rectangle (from (r1,c1) to (r2,c2)) to '0'.
9
Step 9: Continue iterating through the matrix until all cells have been processed.
10
Step 10: Return the result list containing the coordinates of all identified farmland groups.

💡

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*(m+n))

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity arises from iterating through the entire matrix and then, for each farmland cell, expanding to find the rectangle's boundaries. The space complexity depends on how many farmland groups there are. The worst case, where most cells are farmland and form many small groups, leads to a space complexity proportional to the number of farmland cells.

📖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)

Instead of iteratively expanding the rectangle, we can use DFS to explore the connected farmland cells from a starting point. This involves recursively visiting adjacent farmland cells and marking them as visited.

Time: O(m*n)Space: O(m*n) in the worst case (for the call stack in the recursive calls).

Trade-offs: DFS can be more intuitive for exploring connected components, but its space complexity can be a concern for very large matrices due to the recursion depth. The iterative solution might be slightly more efficient in practice due to lower overhead.

Breadth-First Search (BFS)

Similar to DFS, BFS can be used to explore the connected farmland cells. It uses a queue to manage the cells to visit.

Time: O(m*n)Space: O(m*n) in the worst case (for the queue).

Trade-offs: BFS and DFS have similar time complexities, but BFS might be preferred if we want to explore the farmland group layer by layer.

✨

Key Insights

1Insight 1: Since farmland groups are rectangular and not adjacent, we can iterate through the matrix and, upon finding a '1', determine the extent of the rectangle it belongs to without worrying about merging adjacent groups.
2Insight 2: Modifying the original input matrix 'land' by setting visited farmland to '0' avoids double-counting farmland and simplifies the process of finding distinct groups.
3Insight 3: The problem inherently involves nested loops to traverse the matrix, and the rectangle detection adds further complexity. Efficiently determining the boundaries of the rectangle is important for performance.

📚

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 matrix (land). The code implicitly handles this correctly because the outer loops won't execute, and an empty list will be returned.
Edge case 2: Matrix with all 0s (no farmland). The code handles this correctly because the 'if land[r1][c1] == 1' condition will always be false, and an empty list will be returned.
Edge case 3: Matrix with a single farmland cell. The code correctly identifies the top-left and bottom-right as the same cell.
Edge case 4: Matrix with a single large farmland group filling the entire matrix. The code efficiently identifies the top-left and bottom-right corners without unnecessary iterations.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to mark visited farmland cells. This can lead to infinite loops or incorrect results as the same farmland group might be processed multiple times.

✗Common mistake 2: Incorrectly determining the bottom-right corner of the rectangle. The while loop conditions must be carefully defined to ensure that the expansion stops at the correct boundaries.

✗Common mistake 3: Not handling edge cases properly, such as an empty matrix or a matrix with only 0s. This can lead to runtime errors or incorrect output.

✗Common mistake 4: In C implementation, forgetting to allocate and free memory properly. This can lead to memory leaks and crashes.

✅

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 rectangular regions of interest in satellite imagery or medical scans. Farmland detection can be used for agricultural monitoring and resource management.

🏦Real-world scenario 2: Data mining - Discovering rectangular patterns in datasets, such as identifying customer segments with similar characteristics based on multiple attributes.

🎮Real-world scenario 3: Robotics - Navigation and path planning for robots in environments with rectangular obstacles or regions of interest.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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 13, 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.

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 "Find All Groups of Farmland"?

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.

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