<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 noOfIslands(vector<vector<int>> grid)
    {
        int n = grid.size();
        int m = grid[0].size();
        int noOfIs = 0;
        vector<vector<int>> dir{{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(grid[i][j] == 1) // land
                {
                    noOfIs++;
                    queue<pair<int, int>> q;
                    q.push({i, j});
                    while(!q.empty())
                    {
                        pair<int, int> nd = q.front();
                        q.pop();
                        int ci = nd.first, cj = nd.second;
                        if(grid[ci][cj] == 0) // water
                            continue;
                        grid[ci][cj] = 0;
                        for(vector<int> &d: dir)
                        {
                            int ni = ci + d[0], nj = cj + d[1];
                            if(ni < 0 || ni >= n || nj < 0 || nj >= m)
                                continue;
                            if(grid[ni][nj] == 1) // land
                                q.push({ni, nj});
                        }
                    }
                }
            }
        }
        return noOfIs;
    }

public:
    int minDays(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        if(noOfIslands(grid) != 1)
            return 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = 0; j < m; j++)
            {
                if(grid[i][j] == 0)
                    continue;
                grid[i][j] = 0;
                if(noOfIslands(grid) != 1)
                    return 1;
                grid[i][j] = 1;
            }
        }
        return 2;
    }
};
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