<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 {
    int rnbrs[4] = {-1,0,0,1};
    int cnbrs[4] = {0,1,-1,0};
    int m,n;
public:
    int minDays(vector<vector<int>>& grid) {
        m = grid.size();
        n = grid[0].size();        
        std::vector<std::pair<int,int>> coords;                
        if (count_conn_comps(grid, coords) != 1) return 0;
        reset(grid);
        for (int i = 0; i < coords.size(); ++i) {
            std::vector<std::pair<int,int>> temp_coords;
            grid[coords[i].first][coords[i].second] = 0;
            if (count_conn_comps(grid, temp_coords) != 1) return 1;
            reset(grid);
            grid[coords[i].first][coords[i].second] = 1;
        }
        return 2;
    }
private:
    int count_conn_comps(vector<vector<int>>& grid, std::vector<std::pair<int,int>>& coords) {
        int conn_comp = 0;
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (grid[i][j] == 1) {
                    ++conn_comp;
                    dfs(i, j, grid, coords);
                }
        return conn_comp;
    }
    void reset(vector<vector<int>>& grid) {
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (grid[i][j] == 2) grid[i][j] = 1;
    }
    void dfs(int x, int y, vector<vector<int>>& grid, std::vector<std::pair<int,int>>& coords) {
        grid[x][y] = 2;
        coords.emplace_back(x, y);
        for (int k = 0; k < 4; ++k) {
            int next_x = x + rnbrs[k];
            int next_y = y + cnbrs[k];
            if (next_x < 0 || next_x == m || next_y < 0 || next_y == n || grid[next_x][next_y] == 0)
                continue;
            if (grid[next_x][next_y] == 1)
                dfs(next_x, next_y, grid, coords);
        }                
    }
};
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