<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:
    int dx[4] = {-1, 0, 1, 0};
    int dy[4] = {0, 1, 0, -1};
    
    void bfs(int row, int col, vector<vector<int>>& land, vector<int>& temp, 
             vector<pair<int, int>>& v, vector<vector<int>>& visited) {
        visited[row][col] = true; 
        v.push_back({row, col}); 
        for (int k = 0; k < 4; k++) {
            int nr = row + dx[k];
            int nc = col + dy[k];
            if (nr >= 0 && nr < land.size() && nc >= 0 && nc < land[0].size() && 
                land[nr][nc] == 1 && !visited[nr][nc]) {
                bfs(nr, nc, land, temp, v, visited);
            }
        }
    }
    
    vector<vector<int>> findFarmland(vector<vector<int>>& land) {
        vector<vector<int>> result;
        vector<pair<int, int>> v;
        int m = land.size();
        int n = land[0].size();
        vector<vector<int>> visited(m, vector<int>(n, false));
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (land[i][j] == 1 && !visited[i][j]) {
                    vector<int> temp;
                    temp.push_back(i);  // Add starting row
                    temp.push_back(j);  // Add starting column
                    bfs(i, j, land, temp, v, visited);  
                    // Sort v to get the bottom-right corner of the farmland
                    sort(v.begin(), v.end());
                    if (!v.empty()) {
                        auto [er, ec] = v.back();  
                        temp.push_back(er);
                        temp.push_back(ec);
                    }else{
                        temp.push_back(i);
                        temp.push_back(j);
                    }
                    v.clear();  // Clear for the next group of farmland
                    result.push_back(temp);
                }
            }
        }
        
        return result;
    }
};

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