<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 {
using PII = pair<int ,int>;
public:

    int island(vector<vector<int>>& grid, vector<int>& dirX, vector<int>& dirY)
    {
        int n = grid.size(), m = grid[0].size();
        queue<PII> q;
        int islandNum = 0;
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(grid[i][j]==0 || grid[i][j]==-1) continue;
                if(grid[i][j]==1)
                {
                    q.push({i, j});
                    grid[i][j] = -1;
                    islandNum++;
                } 
                while(!q.empty())
                {
                    PII curr = q.front();
                    q.pop();
                    for(int k=0; k<4; k++)
                    {
                        int newX = curr.second + dirX[k];
                        int newY = curr.first + dirY[k];
                        if(newX>=0 && newX<m && newY>=0 && newY<n && grid[newY][newX]==1)
                        {
                            q.push({newY, newX});
                            grid[newY][newX] = -1;
                        }
                    }
                }
            }
        }
        return islandNum;
    }
    void resetGrid(vector<vector<int>>& grid)
    {
        int n = grid.size(), m = grid[0].size();
        for(int k=0; k<n; k++)
        {
            for(int p=0; p<m; p++)
            {
                if(grid[k][p] == -1)
                {
                    grid[k][p] = 1;
                }
            }
        }
    }
    int minDays(vector<vector<int>>& grid) 
    {
        vector<int> dirX = {-1, 0, 0, 1};
        vector<int> dirY = {0, 1, -1, 0};
        int islandNums = island(grid, dirX, dirY);
        resetGrid(grid);
        if(islandNums>1 || islandNums==0) return 0;
        int n = grid.size(), m = grid[0].size();
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<m; j++)
            {
                if(!grid[i][j]) continue;
                grid[i][j] = 0;
                islandNums = island(grid, dirX, dirY);
                if(islandNums>1 || islandNums==0) return 1;
                resetGrid(grid);
                grid[i][j] = 1;
            }
        }
        return 2;
    }
};
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