<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:
    vector<vector<int>> dir = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    vector<vector<bool>> vis;
    vector<vector<int>> matrix;
    void bfs(int r, int c) {
        queue<pair<int, int>> q;
        vis[r][c] = true;
        q.push({r, c});
        while (q.size() > 0) {
            auto cell = q.front();
            q.pop();
            int x = cell.first;
            int y = cell.second;
            for (int i = 0; i < 4; i++) {
                int cx = x + dir[i][0];
                int cy = y + dir[i][1];
                if (cx >= 0 and cy >= 0 and cx < matrix.size() and
                    cy < matrix[0].size() and vis[cx][cy] == false and
                    matrix[cx][cy] == 1) {
                    vis[cx][cy] = true;
                    q.push({cx, cy});
                }
            }
        }
    }
    int minDays(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        vis.clear();
        vis.resize(m, vector<bool>(n, false));
        matrix = grid;
        int islands = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (vis[i][j] == false and grid[i][j] == 1) {
                    islands++;
                    bfs(i, j);
                }
            }
        }
        if (islands == 0 or islands >= 2)
            return 0;

        // islands==1
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (matrix[i][j] == 1) {
                    matrix[i][j] = 0;
                    vis.clear();
                    vis.resize(m, vector<bool>(n, false));
                    int island = 0;
                    for (int p = 0; p < m; p++) {
                        for (int q = 0; q < n; q++) {
                            if (vis[p][q] == false and matrix[p][q] == 1) {
                                island++;
                                bfs(p, q);
                            }
                        }
                    }
                    if (island >= 2 or island==0)
                        return 1;
                    matrix[i][j] = 1;
                }
            }
        }
        return 2;
    }
};
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