<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 i,int j){
        int m=grid.size();
        int n=grid[0].size();
        queue<pair<int,int>> q;
        q.push({i,j});
        visited[i][j]=1;
        while(!q.empty()){
            pair<int,int> p=q.front();
            int row=p.first;
            int col=p.second;
            q.pop();
            if(row+1<m && grid[row+1][col]==1 && visited[row+1][col]==0){
                q.push({row+1,col});
                visited[row+1][col]=1;
            }
            if(row-1>=0 && grid[row-1][col]==1 && visited[row-1][col]==0){
                q.push({row-1,col});
                visited[row-1][col]=1;
            }
            if(col+1<n && grid[row][col+1]==1 && visited[row][col+1]==0){
                q.push({row,col+1});
                visited[row][col+1]=1;
            }
            if(col-1>=0 && grid[row][col-1]==1 && visited[row][col-1]==0){
                q.push({row,col-1});
                visited[row][col-1]=1;
            }

        }
    }
    int numIslands(vector<vector<int>>& grid){
        int m=grid.size();
        int n=grid[0].size();
        int cnt=0;
        vector<vector<int>> visited(m,vector<int> (n,0));
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(visited[i][j]==0 && grid[i][j]==1){
                    cnt++;
                    bfs(grid,visited,i,j);
                }
            }
        }
        return cnt;
    } 

    int minDays(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int islands = numIslands(grid);

        //Grid already disconnected
        if (islands > 1 or islands == 0) {
            return 0;
        } else {
			// check for 1 ans
            for (int i = 0 ; i < m; i ++) {
                for (int j = 0; j < n; j++) {
                    if (grid[i][j] == 1) {
                        grid[i][j] = 0; //visited
						
                        islands = numIslands(grid);
						
                        grid[i][j] = 1; //unvisited
                        if (islands > 1 or islands == 0)
                            return 1;
                    }

                }
            }
        }

        return 2; //It's always possible to break an island with 2 moves
    }
};
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