<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:
    void bfs(vector<vector<int>>& grid, vector<vector<int>>& visited, int r, int c) {
        queue<pair<int,int>> q;
        q.push({r,c});
        visited[r][c] = 1;
        int rows = grid.size();
        int cols = grid[0].size();
        
        vector<int> dr = {-1, 1, 0, 0};
        vector<int> dc = {0, 0, -1, 1};
        
        while(!q.empty()) {
            int cr = q.front().first;
            int cc = q.front().second;
            q.pop();
            
            for(int i = 0; i < 4; i++) {
                int nr = cr + dr[i];
                int nc = cc + dc[i];
                
                if(nr >= 0 && nr < rows && nc >= 0 && nc < cols && grid[nr][nc] == 1 && visited[nr][nc] == 0) {
                    visited[nr][nc] = 1;
                    q.push({nr, nc});
                }
            }
        }
    }

    int numberOfIslands(vector<vector<int>>& grid) {
        int count = 0;
        int rows = grid.size();
        int cols = grid[0].size();
        vector<vector<int>> visited(rows, vector<int>(cols, 0));
        
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                if(grid[i][j] == 1 && visited[i][j] == 0) {
                    count++;
                    bfs(grid, visited, i, j);
                }
            }
        }
        return count;
    }

    int minDays(vector<vector<int>>& grid) {
        int initialIslands = numberOfIslands(grid);
        if (initialIslands != 1) {
            return 0;
        }
        
        // Check if removing any single land cell disconnects the island
        int rows = grid.size();
        int cols = grid[0].size();
        
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 1) {
                    grid[i][j] = 0;
                    int newIslands = numberOfIslands(grid);
                    if (newIslands > 1 || newIslands == 0) {
                        return 1;
                    }
                    grid[i][j] = 1;
                }
            }
        }
        
        // If removing a single cell doesn't disconnect, return 2
        return 2;
    }
};

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