<p>Given a 2D&nbsp;<code>grid</code> consists of <code>0s</code> (land)&nbsp;and <code>1s</code> (water).&nbsp; An <em>island</em> is a maximal 4-directionally connected group of <code><font face="monospace">0</font>s</code> and a <em>closed island</em>&nbsp;is an island <strong>totally</strong>&nbsp;(all left, top, right, bottom) surrounded by <code>1s.</code></p>

<p>Return the number of <em>closed islands</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_3_1610.png" style="width: 240px; height: 120px;" /></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 
Islands in gray are closed because they are completely surrounded by water (group of 1s).</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2019/10/31/sample_4_1610.png" style="width: 160px; height: 80px;" /></p>

<pre>
<strong>Input:</strong> grid = [[0,0,1,0,0],[0,1,0,1,0],[0,1,1,1,0]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1,1,1],
&nbsp;              [1,0,0,0,0,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,1,0,1,0,1],
&nbsp;              [1,0,1,1,1,0,1],
&nbsp;              [1,0,0,0,0,0,1],
               [1,1,1,1,1,1,1]]
<strong>Output:</strong> 2
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= grid.length, grid[0].length &lt;= 100</code></li>
	<li><code>0 &lt;= grid[i][j] &lt;=1</code></li>
</ul>

class Solution {
public:
    int numIslands = 0; 
    int closedIsland(vector<vector<int>>& grid) {
        //create a duplicate of grid where we mark all the visited nodes via a bool, while loop to iterate through everything 
        //while everything isn't visited run dfs
        //if we haven't visited and we hit closed island on all sides then we add one more to the count of closed islands 

        //vector<vector<bool>> grid_; 

        vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), false)); 
        
       for(int i = 0; i < grid.size(); i++) { 
        for(int j = 0; j < grid[i].size(); j++) { 
            if(grid[i][j] == 0 && visited[i][j] == false) { 
                queue<pair<int, int>> queue_; 
                queue_.push({i, j}); 
                if(bfs(queue_, grid, visited)) {
                    numIslands++; 
                }
                
            }
        }
       }
        return numIslands; 
    }
    
    bool bfs(queue<pair<int, int>> queue_, vector<vector<int>>& grid, vector<vector<bool>>& visited) { 
        int i = queue_.front().first; 
        int j = queue_.front().second; 
        bool isClosed = true; 
        
        while(!queue_.empty()) { 
            i = queue_.front().first; 
            j = queue_.front().second; 
            queue_.pop();
            
            if(i == 0 || j == 0 || i == grid.size() - 1 || j == grid[0].size() - 1) {
                isClosed = false; 
            }
            
            if(i + 1 < grid.size() && grid[i+1][j] == 0 && visited[i+1][j] == false) { 
                queue_.push({i+1, j});
                visited[i+1][j] = true; 
            }
            if(i - 1 >= 0 && grid[i-1][j] == 0 && visited[i-1][j] == false) { 
                queue_.push({i-1, j});
                visited[i-1][j] = true; 
            }
            if(j - 1 >= 0 && grid[i][j-1] == 0 && visited[i][j-1] == false) { 
                queue_.push({i, j-1}); 
                visited[i][j-1] = true; 
            }
            if(j + 1 < grid[0].size() && grid[i][j+1] == 0 && visited[i][j+1] == false) { 
                queue_.push({i, j+1}); 
                visited[i][j+1] = true; 
            }
            

        }
        return isClosed; 

    }

};
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