<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 {
  private:
    void bfs(int r, int c, vector<vector<int>>& grid, vector<vector<bool>>& visited){
      vector<vector<int>> directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
      queue<pair<int, int>> bfsQ;
      bfsQ.push(make_pair(r, c));
      visited[r][c] = true;
      while(!bfsQ.empty()){
        pair<int, int> n = bfsQ.front();
        bfsQ.pop();
        for(size_t i = 0; i < directions.size(); i++){
          int targetR = n.first + directions[i][0];
          int targetC = n.second + directions[i][1];
          if(targetR >= 0 && targetR < grid.size() && targetC >= 0 && targetC < grid[0].size() && 
            grid[targetR][targetC] == 1 && visited[targetR][targetC] == false){
              bfsQ.push(make_pair(targetR, targetC));
              visited[targetR][targetC] = true;
          }
        }
      }
    }
    int countIslands(vector<vector<int>>& grid){
      int numIslands = 0;
      vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), false));
      for(size_t r = 0; r < grid.size(); r++){
        for(size_t c = 0; c < grid[0].size(); c++){
          if(grid[r][c] == 1 && !visited[r][c]){
            bfs(r, c, grid, visited);
            numIslands++;
          }
        }
      }
      return numIslands;
    }
  public:
    int minDays(vector<vector<int>>& grid) {
      if(countIslands(grid) != 1){
        return 0;
      }
      for(size_t r = 0; r < grid.size(); r++){
        for(size_t c = 0; c < grid[0].size(); c++){
          if(grid[r][c] == 1){
            //try to change to water
            grid[r][c] = 0;
            //successfully disconnected graph by disconnecting this cell
            if(countIslands(grid) != 1){
              return 1;
            }
            //revert change
            grid[r][c] = 1;
          }
        }
      }
      return 2;
    }
};
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