<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) {
        m = grid.size();
        n = grid[0].size();

        if (!isConnected(grid)) return 0; // Already disconnected

        // Try to disconnect by removing one land cell
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    grid[i][j] = 0;
                    if (!isConnected(grid)) return 1;
                    grid[i][j] = 1; // Restore the grid
                }
            }
        }

        return 2; // Requires removing at least two cells
    }

private:
    int m, n;
    vector<int> direction{-1, 0, 1, 0, -1};

    bool isConnected(vector<vector<int>>& grid) {
        int components = 0;
        vector<vector<bool>> visited(m, vector<bool>(n, false));

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    components++;
                    if (components > 1) return false; // More than one island
                    bfs(grid, i, j, visited);
                }
            }
        }

        return components == 1;
    }

    void bfs(vector<vector<int>>& grid, int i, int j, vector<vector<bool>>& visited) {
        queue<pair<int, int>> q;
        q.push({i, j});
        visited[i][j] = true;

        while (!q.empty()) {
            auto [x, y] = q.front();
            q.pop();

            for (int d = 0; d < 4; d++) {
                int nx = x + direction[d];
                int ny = y + direction[d + 1];

                if (nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] == 1 && !visited[nx][ny]) {
                    visited[nx][ny] = true;
                    q.push({nx, ny});
                }
            }
        }
    }
};
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