<p>You are given an <code>m x n</code> binary grid <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water. An <strong>island</strong> is a maximal <strong>4-directionally</strong> (horizontal or vertical) connected group of <code>1</code>&#39;s.</p>

<p>The grid is said to be <strong>connected</strong> if we have <strong>exactly one island</strong>, otherwise is said <strong>disconnected</strong>.</p>

<p>In one day, we are allowed to change <strong>any </strong>single land cell <code>(1)</code> into a water cell <code>(0)</code>.</p>

<p>Return <em>the minimum number of days to disconnect the grid</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land1.jpg" style="width: 500px; height: 169px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]

<strong>Output:</strong> 2
<strong>Explanation:</strong> We need at least 2 days to get a disconnected grid.
Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land2.jpg" style="width: 404px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [[1,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Grid of full water is also disconnected ([[1,1]] -&gt; [[0,0]]), 0 islands.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:

    int dr[4] = {-1, 0, 0, 1};
    int dc[4] = {0, 1, -1, 0};
    int n, m;
    int vis[31][31];

    void dfs(int row, int col, vector<vector<int>> &grid, vector<pair<int, int>> &v) {
        vis[row][col] = 1;
        v.push_back({row, col});

        for (int i = 0; i < 4; i++) {
            int r = dr[i] + row;
            int c = dc[i] + col;
            if (r >= 0 && c >= 0 && r < n && c < m && !vis[r][c] && grid[r][c]) {
                dfs(r, c, grid, v);
            }
        }
    }

    int minDays(vector<vector<int>>& grid) {
        n = grid.size();
        m = grid[0].size();
        memset(vis, 0, sizeof(vis));
        int scc = 0;
        vector<pair<int, int>> v;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (!vis[i][j] && grid[i][j]) {
                    scc++;
                    dfs(i, j, grid, v);
                }
            }
        }

        if (scc == 0 || scc > 1) return 0;

        for (auto it : v) {
            grid[it.first][it.second] = 0;
            scc = 0;
            memset(vis, 0, sizeof(vis));
            vector<pair<int, int>> tmp_v; 
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if (!vis[i][j] && grid[i][j]) {
                        scc++;
                        dfs(i, j, grid, tmp_v);
                    }
                }
            }
            grid[it.first][it.second] = 1;
            if (scc > 1 || scc==0) return 1;
        }
        return 2;
    }

};
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