<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 count = 0;
    int direction[4][2] = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
    void dfs(vector<vector<int>>& grid, int i, int j, int& curCount,
             vector<vector<bool>>& visited) {
        visited[i][j] = 1;
        curCount++;
        for (int k = 0; k < 4; k++) {
            int x = i + direction[k][0];
            int y = j + direction[k][1];

            if (x >= 0 && x < grid.size() && y >= 0 && y < grid[0].size())
            {
                if(grid[x][y] && !visited[x][y]) {
                dfs(grid, x, y, curCount, visited);}
            }
        }
        return;
    }

    int minDays(vector<vector<int>>& grid) {
        vector<vector<bool>> visited(31, vector<bool>(31, 0));
        count = 0;
        int x, y;
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j]) {
                    x = i;
                    y = j;
                    count++;
                }
            }
        }
        if (count == 0)
            return 0;
        int curCount = 0;
        dfs(grid, x, y, curCount, visited);
        if (curCount != count) {
            return 0;
        }
        if (curCount == 2) {
            return 2;
        }
        if (curCount == 1) {
            return 1;
        }
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j]) {
                    grid[i][j] = 0;
                    vector<vector<bool>> visited1(31, vector<bool>(31, 0));

                    curCount = 0;
                            
                    for (int k = 0; k < 4; k++) {
                        x = i + direction[k][0];
                        y = j + direction[k][1];
                        if (x >= 0 && x < grid.size() && y >= 0 &&
                            y < grid[0].size() && grid[x][y]) {
                            dfs(grid, x, y, curCount, visited1);
                            break;
                        }
                    }
                    grid[i][j] = 1;

                    if (curCount != count - 1) {
                        return 1;
                    }
                }
            }
        }
        return 2;
    }
};
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