<p>You are given a <strong>0-indexed</strong> <code>m x n</code> binary matrix <code>land</code> where a <code>0</code> represents a hectare of forested land and a <code>1</code> represents a hectare of farmland.</p>

<p>To keep the land organized, there are designated rectangular areas of hectares that consist <strong>entirely</strong> of farmland. These rectangular areas are called <strong>groups</strong>. No two groups are adjacent, meaning farmland in one group is <strong>not</strong> four-directionally adjacent to another farmland in a different group.</p>

<p><code>land</code> can be represented by a coordinate system where the top left corner of <code>land</code> is <code>(0, 0)</code> and the bottom right corner of <code>land</code> is <code>(m-1, n-1)</code>. Find the coordinates of the top left and bottom right corner of each <strong>group</strong> of farmland. A <strong>group</strong> of farmland with a top left corner at <code>(r<sub>1</sub>, c<sub>1</sub>)</code> and a bottom right corner at <code>(r<sub>2</sub>, c<sub>2</sub>)</code> is represented by the 4-length array <code>[r<sub>1</sub>, c<sub>1</sub>, r<sub>2</sub>, c<sub>2</sub>].</code></p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<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;" />
<pre>
<strong>Input:</strong> land = [[1,0,0],[0,1,1],[0,1,1]]
<strong>Output:</strong> [[0,0,0,0],[1,1,2,2]]
<strong>Explanation:</strong>
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].
</pre>

<p><strong class="example">Example 2:</strong></p>
<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;" />
<pre>
<strong>Input:</strong> land = [[1,1],[1,1]]
<strong>Output:</strong> [[0,0,1,1]]
<strong>Explanation:</strong>
The first group has a top left corner at land[0][0] and a bottom right corner at land[1][1].
</pre>

<p><strong class="example">Example 3:</strong></p>
<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;" />
<pre>
<strong>Input:</strong> land = [[0]]
<strong>Output:</strong> []
<strong>Explanation:</strong>
There are no groups of farmland.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == land.length</code></li>
	<li><code>n == land[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>land</code> consists of only <code>0</code>&#39;s and <code>1</code>&#39;s.</li>
	<li>Groups of farmland are <strong>rectangular</strong> in shape.</li>
</ul>

class Solution:
    def findFarmland(self, land: List[List[int]]) -> List[List[int]]:
        # Edge case: empty land
        if len(land) == 0 or len(land[0]) == 0:
            return []

        # Directions for moving: down, up, left, right
        directions = [(1, 0), (-1, 0), (0, -1), (0, 1)]

        # DFS function to explore the farmland and mark it as visited
        def dfs(row, column, current_indexs):

            # Base case: check if out of bounds or if already visited
            if row < 0 or row >= len(land) or column < 0 or column >= len(land[0]):
                return current_indexs

            # Check if not visited before or if it's not farmland (0)
            if land[row][column] == "#" or land[row][column] == 0:
                return current_indexs

            # Mark the current cell as visited
            land[row][column] = "#"  # Mark this cell as visited by changing it to '#'

            # Update the bottom-right corner (largest row and column)
            current_indexs[2] = max(current_indexs[2], row)
            current_indexs[3] = max(current_indexs[3], column)

            # Explore the 4 directions (expand the farmland area)
            for direction in directions:
                row_add, column_add = direction
                current_indexs = dfs(row + row_add, column + column_add, current_indexs)

            return current_indexs

        # Main loop to iterate through the land
        output_all_lands = []

        for row in range(len(land)):
            for column in range(len(land[0])):
                if land[row][column] == 1:  # If this is a new farmland (unvisited)
                    # Start DFS with the current land cell and initialize top-left and bottom-right
                    cur_index = dfs(row, column, [row, column, row, column])
                    output_all_lands.append(cur_index)

        return output_all_lands

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