Introduction

In this blog post, we will explore how to solve the LeetCode problem "1992" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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.

Explanation

Here's a breakdown of the solution:

• Iterate and Identify: Scan the matrix from top-left to bottom-right. When a '1' (farmland) is encountered, it marks the top-left corner of a new group.
• Expand and Mark: From the top-left, expand the group to find its bottom-right corner. As we determine the group's area, mark the farmland cells within the group as visited (e.g., by changing them to '0') to avoid processing them again.

• Collect and Return: Store the top-left and bottom-right coordinates of each identified group in the result array and return it.

• Runtime Complexity: O(m*n), where m is the number of rows and n is the number of columns in the matrix.

• Storage Complexity: O(1), excluding the output array, as we modify the input matrix in place.

Code

    def findFarmland(land):
    """
    Finds the coordinates of the top-left and bottom-right corner of each group of farmland.

    Args:
        land: A 0-indexed m x n binary matrix representing farmland (1) and forested land (0).

    Returns:
        A 2D array containing the coordinates of each group of farmland.
    """

    m = len(land)
    n = len(land[0])
    result = []

    for r in range(m):
        for c in range(n):
            if land[r][c] == 1:
                # Found the top-left corner of a group
                r1, c1 = r, c
                r2, c2 = r, c

                # Find the bottom-right corner
                while r2 + 1 < m and land[r2 + 1][c] == 1:
                    r2 += 1
                while c2 + 1 < n and land[r][c2 + 1] == 1:
                    c2 += 1

                # Extend r2 and c2 to cover all 1s in the group
                temp_r2 = r
                while temp_r2 <= r2:
                    temp_c2 = c
                    while temp_c2 <= c2:
                        if land[temp_r2][temp_c2] != 1:
                            break
                        temp_c2+=1
                    temp_r2+=1

                result.append([r1, c1, r2, c2])

                # Mark the farmland as visited (set to 0)
                for i in range(r1, r2 + 1):
                    for j in range(c1, c2 + 1):
                        land[i][j] = 0

    return result