<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 subnum;
    int n, m;
public:
    int minDays(vector<vector<int>>& grid) {
        // We can disconnect the grid within at most 2 days.
        // remove one grid is enough.
        n = grid.size();
        m = grid[0].size();
        // find how many component
        vector<vector<bool>> trace(n, vector<bool>(m, false));
        bool find = false;
        int minConnect = 4;
        int num = 0;
        // also check if any gird only connect one component;
        for(int i=0; i<n; ++i){
            for(int j=0; j<m; ++j){
                if(grid[i][j] == 1 && trace[i][j] == false && find == true)
                    return 0;   // already disconnect
                else if(grid[i][j] == 1 && trace[i][j] == false){
                    find = true;
                    // do BFS
                    queue<pair<int, int>> q;
                    q.push({i,j});
                    trace[i][j] = true;
                    while(!q.empty()){
                        auto [x,y] = q.front();
                        q.pop();
                        ++num;
                        int count = 0;
                        if(x>0 && grid[x-1][y]){
                            ++count;
                            if(trace[x-1][y] == false){
                                q.push({x-1, y});
                                trace[x-1][y] = true;
                            }
                        }
                        if(y>0 && grid[x][y-1]){
                            ++count;
                            if(trace[x][y-1] == false){
                                q.push({x, y-1});
                                trace[x][y-1] = true;
                            }
                        }
                        if(x<n-1 && grid[x+1][y]){
                            ++count;
                            if(trace[x+1][y] == false){
                                q.push({x+1, y});
                                trace[x+1][y] = true;
                            }
                        }
                        if(y<m-1 && grid[x][y+1]){
                            ++count;
                            if(trace[x][y+1] == false){
                                q.push({x, y+1});
                                trace[x][y+1] = true;
                            }
                        }
                        if(count < minConnect)
                            minConnect = count;
                    }
                }
            }
        }
        if(num <= 2)
            return num;
        if(minConnect == 1)
            return 1;
        // remove each grid to check if we can only remove one gird
        for(int i=0; i<n; ++i){
            for(int j=0; j<m; ++j){
                if(grid[i][j] == 1){
                    grid[i][j] = 0;
                    subnum = 0;
                    if(i>0 && grid[i-1][j]){
                        runBFS(grid, i-1, j);
                    } else if(j>0 && grid[i][j-1]){
                        runBFS(grid, i, j-1);
                    } else if(i < n-1 && grid[i+1][j]){
                        runBFS(grid, i+1, j);
                    } else {
                        runBFS(grid, i, j+1);
                    }
                    if(num-1 != subnum)
                        return 1;
                    grid[i][j] = 1;
                }
            }
        }
        return 2;
    }
    void runBFS(vector<vector<int>>& grid, int i, int j){
        vector<vector<bool>> trace(n, vector<bool>(m, false));
        queue<pair<int, int>> q;
        q.push({i, j});
        trace[i][j] = true;
        while(!q.empty()){
            auto [x,y] = q.front();
            q.pop();
            ++subnum;
            if(x>0 && grid[x-1][y] && trace[x-1][y] == false){
                q.push({x-1, y});
                trace[x-1][y] = true;
            }
            if(y>0 && grid[x][y-1] && trace[x][y-1] == false){
                q.push({x, y-1});
                trace[x][y-1] = true;
            }
            if(x<n-1 && grid[x+1][y] && trace[x+1][y] == false){
                q.push({x+1, y});
                trace[x+1][y] = true;
            }
            if(y<m-1 && grid[x][y+1] && trace[x][y+1] == false){
                q.push({x, y+1});
                trace[x][y+1] = true;
            }
        }
    }
};
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