<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 {
public:
    vector<vector<int>> findFarmland(vector<vector<int>>& land) {
        vector<vector<bool>> visited(land.size(), vector<bool>(land[0].size(), false));
        vector<vector<int>> ans;
        for (int row1 = 0; row1 < land.size(); row1++) {
            for (int col1 = 0; col1 < land[0].size(); col1++) {
                if (land[row1][col1] && !visited[row1][col1]) {
                    int row2 = 0, col2 = 0;
                    DFS(land, visited, row1, col1, row2, col2);
                    ans.push_back({row1, col1, row2, col2});
                }
            }
        }
        
        return ans;
    }

private:
    // The four directions in which traversal will be done.
    int dirs[4][2] = {{-1, 0}, {0, -1}, {0, 1}, {1, 0}};
    // Returns true if the coordinate is within the boundary of the matrix.
    bool isWithinFarm(int x, int y, int N, int M) {
        return x >= 0 && x < N && y >= 0 && y < M;
    }
    
    void DFS(vector<vector<int>>& land, vector<vector<bool>>& visited, int x, int y, int& row2,
        int& col2) {
        visited[x][y] = true;
        // Maximum x and y for the bottom right cell.
        row2 = max(row2, x); col2 = max(col2, y);
        
        for (auto dir : dirs) {
            // Neighbor cell coordinates.
            int newX = x + dir[0], newY = y + dir[1];
            // If the neighbor is within the matrix and is a farmland cell and is not visited yet.
            if (isWithinFarm(newX, newY, land.size(), land[0].size()) && !visited[newX][newY]
                    && land[newX][newY]) {
                DFS(land, visited, newX, newY, row2, col2);
            }
        }
    }    
};
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