<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 minDays(vector<vector<int>>& grid) {
        int inum = islandCount(grid);

        if (inum == 0 || inum > 1) return 0;
        int xs = grid.size(), ys = grid[0].size();
        for (int i = 0; i < xs; ++i) {
            for (int j = 0; j < ys; ++j) {
                if (grid[i][j] != 0) {
                    vector<vector<int>> g = grid;
                    g[i][j] = 0;
                    if (islandCount(g) != 1) return 1;
                }
            }
        }
        
        return 2;
    }

    int islandCount(vector<vector<int>>& grid) {
        int count = 0;
        vector<vector<int>> g = grid;
        int xs = g.size(), ys = g[0].size();
        for (int i = 0; i < xs; ++i) {
            for (int j = 0; j < ys; ++j) {
                if (g[i][j] != 0) {
                    count++;
                    gBFS(g, i, j);
                }
            }
        }
        return count;
    }

    void gBFS(vector<vector<int>>& g, int x, int y) {
        g[x][y] = 0;
        int xmax = g.size()-1, ymax = g[0].size()-1;
        if (x > 0 && g[x-1][y] != 0) gBFS(g, x-1, y);
        if (x < xmax && g[x+1][y] != 0) gBFS(g, x+1, y);
        if (y > 0 && g[x][y-1] != 0) gBFS(g, x, y-1);
        if (y < ymax && g[x][y+1] != 0) gBFS(g, x, y+1);
    }
};
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