LeetCode #1992 — MEDIUM

Find All Groups of Farmland

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 corner of land is (0, 0) and the bottom right corner of land is (m-1, n-1). Find the coordinates of the top left and bottom right corner of each group of farmland. A group of farmland with a top left corner at (r₁, c₁) and a bottom right corner at (r₂, c₂) is represented by the 4-length array [r₁, c₁, r₂, c₂].

Return a 2D array containing the 4-length arrays described above for each group of farmland in land. If there are no groups of farmland, return an empty array. You may return the answer in any order.

Example 1:

Input: land = [[1,0,0],[0,1,1],[0,1,1]]
Output: [[0,0,0,0],[1,1,2,2]]
Explanation:
The first group has a top left corner at land[0][0] and a bottom right corner at land[0][0].
The second group has a top left corner at land[1][1] and a bottom right corner at land[2][2].

Example 2:

Input: land = [[1,1],[1,1]]
Output: [[0,0,1,1]]
Explanation:
The first group has a top left corner at land[0][0] and a bottom right corner at land[1][1].

Example 3:

Input: land = [[0]]
Output: []
Explanation:
There are no groups of farmland.

Constraints:

m == land.length
n == land[i].length
1 <= m, n <= 300
land consists of only 0's and 1's.
Groups of farmland are rectangular in shape.

Step 01

Brute Force Baseline

Problem summary: 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 corner of land is (0, 0) and the bottom right corner of land is (m-1, n-1). Find the coordinates of the top left and bottom right corner of each group of farmland. A group of farmland with a top left corner at (r1, c1) and a bottom right corner at (r2, c2) is represented by the 4-length array [r1, c1, r2, c2]. Return a 2D array containing the 4-length arrays described above for each group of

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[[1,0,0],[0,1,1],[0,1,1]]

Example 2

[[1,1],[1,1]]

Example 3

[[0]]

Core Insight

What unlocks the optimal approach

Since every group of farmland is rectangular, the top left corner of each group will have the smallest x-coordinate and y-coordinate of any farmland in the group.
Similarly, the bottom right corner of each group will have the largest x-coordinate and y-coordinate of any farmland in the group.
Use DFS to traverse through different groups of farmlands and keep track of the smallest and largest x-coordinate and y-coordinates you have seen in each group.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1992: Find All Groups of Farmland
class Solution {
    public int[][] findFarmland(int[][] land) {
        List<int[]> ans = new ArrayList<>();
        int m = land.length;
        int n = land[0].length;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (land[i][j] == 0 || (j > 0 && land[i][j - 1] == 1)
                    || (i > 0 && land[i - 1][j] == 1)) {
                    continue;
                }
                int x = i;
                int y = j;
                for (; x + 1 < m && land[x + 1][j] == 1; ++x)
                    ;
                for (; y + 1 < n && land[x][y + 1] == 1; ++y)
                    ;
                ans.add(new int[] {i, j, x, y});
            }
        }
        return ans.toArray(new int[ans.size()][4]);
    }
}

// Accepted solution for LeetCode #1992: Find All Groups of Farmland
func findFarmland(land [][]int) [][]int {
	m, n := len(land), len(land[0])
	var ans [][]int
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if land[i][j] == 0 || (j > 0 && land[i][j-1] == 1) || (i > 0 && land[i-1][j] == 1) {
				continue
			}
			x, y := i, j
			for ; x+1 < m && land[x+1][j] == 1; x++ {
			}
			for ; y+1 < n && land[x][y+1] == 1; y++ {
			}
			ans = append(ans, []int{i, j, x, y})
		}
	}
	return ans
}

# Accepted solution for LeetCode #1992: Find All Groups of Farmland
class Solution:
    def findFarmland(self, land: List[List[int]]) -> List[List[int]]:
        m, n = len(land), len(land[0])
        ans = []
        for i in range(m):
            for j in range(n):
                if (
                    land[i][j] == 0
                    or (j > 0 and land[i][j - 1] == 1)
                    or (i > 0 and land[i - 1][j] == 1)
                ):
                    continue
                x, y = i, j
                while x + 1 < m and land[x + 1][j] == 1:
                    x += 1
                while y + 1 < n and land[x][y + 1] == 1:
                    y += 1
                ans.append([i, j, x, y])
        return ans

// Accepted solution for LeetCode #1992: Find All Groups of Farmland
/**
 * [1992] Find All Groups of Farmland
 *
 * 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 corner of land is (0, 0) and the bottom right corner of land is (m-1, n-1). Find the coordinates of the top left and bottom right corner of each group of farmland. A group of farmland with a top left corner at (r1, c1) and a bottom right corner at (r2, c2) is represented by the 4-length array [r1, c1, r2, c2].
 * Return a 2D array containing the 4-length arrays described above for each group of farmland in land. If there are no groups of farmland, return an empty array. You may return the answer in any order.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/07/27/screenshot-2021-07-27-at-12-23-15-copy-of-diagram-drawio-diagrams-net.png" style="width: 300px; height: 300px;" />
 * Input: land = [[1,0,0],[0,1,1],[0,1,1]]
 * Output: [[0,0,0,0],[1,1,2,2]]
 * Explanation:
 * The first group has a top left corner at land[0][0] and a bottom right corner at land[0][0].
 * The second group has a top left corner at land[1][1] and a bottom right corner at land[2][2].
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/07/27/screenshot-2021-07-27-at-12-30-26-copy-of-diagram-drawio-diagrams-net.png" style="width: 200px; height: 200px;" />
 * Input: land = [[1,1],[1,1]]
 * Output: [[0,0,1,1]]
 * Explanation:
 * The first group has a top left corner at land[0][0] and a bottom right corner at land[1][1].
 *
 * Example 3:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/07/27/screenshot-2021-07-27-at-12-32-24-copy-of-diagram-drawio-diagrams-net.png" style="width: 100px; height: 100px;" />
 * Input: land = [[0]]
 * Output: []
 * Explanation:
 * There are no groups of farmland.
 *
 *  
 * Constraints:
 *
 * 	m == land.length
 * 	n == land[i].length
 * 	1 <= m, n <= 300
 * 	land consists of only 0's and 1's.
 * 	Groups of farmland are rectangular in shape.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/find-all-groups-of-farmland/
// discuss: https://leetcode.com/problems/find-all-groups-of-farmland/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn find_farmland(land: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
        vec![]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_1992_example_1() {
        let land = vec![vec![1, 0, 0], vec![0, 1, 1], vec![0, 1, 1]];

        let result = vec![vec![0, 0, 0, 0], vec![1, 1, 2, 2]];

        assert_eq!(Solution::find_farmland(land), result);
    }

    #[test]
    #[ignore]
    fn test_1992_example_2() {
        let land = vec![vec![1, 1], vec![1, 1]];

        let result = vec![vec![0, 0, 1, 1]];

        assert_eq!(Solution::find_farmland(land), result);
    }

    #[test]
    #[ignore]
    fn test_1992_example_3() {
        let land = vec![vec![0]];

        let result: Vec<Vec<i32>> = vec![];

        assert_eq!(Solution::find_farmland(land), result);
    }
}

// Accepted solution for LeetCode #1992: Find All Groups of Farmland
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1992: Find All Groups of Farmland
// class Solution {
//     public int[][] findFarmland(int[][] land) {
//         List<int[]> ans = new ArrayList<>();
//         int m = land.length;
//         int n = land[0].length;
//         for (int i = 0; i < m; ++i) {
//             for (int j = 0; j < n; ++j) {
//                 if (land[i][j] == 0 || (j > 0 && land[i][j - 1] == 1)
//                     || (i > 0 && land[i - 1][j] == 1)) {
//                     continue;
//                 }
//                 int x = i;
//                 int y = j;
//                 for (; x + 1 < m && land[x + 1][j] == 1; ++x)
//                     ;
//                 for (; y + 1 < n && land[x][y + 1] == 1; ++y)
//                     ;
//                 ans.add(new int[] {i, j, x, y});
//             }
//         }
//         return ans.toArray(new int[ans.size()][4]);
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.