<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:
    void dfs(int i , int j ,vector<vector<int>>& grid){
        int row = grid.size() ; 
        int col = grid[0].size();
        grid[i][j] = 1 ;

        vector<pair<int,int>>dirs = {{1 , 0} , {0 , 1} , {-1 , 0} , {0,-1}};

        for(pair<int,int>p : dirs){
            int nx = i+p.first ;
            int ny = j+p.second ;

            if(nx>=0 && ny>=0 && nx<row && ny<col && grid[nx][ny] == 0){
                dfs(nx ,ny , grid);
            }
        }
    }
    int closedIsland(vector<vector<int>>& grid) {
        int row = grid.size();
        int col = grid[0].size();

        // Make all Boundary islands one
        /*
        int i = 0 ;
        int j = 0 ;
        while(j<col){
            if(grid[i][j] == 0){
                dfs(i , j , grid);
            }
            j++ ;
        } 

        j = col -1 ;
        i = 1;

        while(i<row){
            if(grid[i][j] == 0){
                dfs(i , j , grid);
            }
            i++ ;
        }
        i = row-1 ;
        j = col-2 ; 
        while(j>=0){
            if(grid[i][j] == 0){
                dfs(i , j , grid);
            }
            j-- ;
        }  
        j = 0 ;
        i = row -2 ;
        while(i>0){
            if(grid[i][j] == 0){
                dfs(i , j , grid);
            }
            i-- ;
        }
        */
        // Another way to call boundary islands
        
        for(int i = 0 ; i<row ; i++){
            for(int j = 0 ; j<col ; j++){
                if(i*j == 0 || i == row-1 || j == col-1){
                    if(grid[i][j]==0){
                        dfs(i , j , grid);
                    }
                }
            }
        } 

        // Call dfs for remaining islands
        int count = 0 ;
        for(int i = 0 ; i<row ; i++){
            for(int j = 0 ; j<col ; j++){
                if(grid[i][j] == 0){
                    count++;
                    dfs(i , j , grid);
                }
            }
        }
        return count ;
    }
};
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