<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:
    def closedIsland(self, grid: List[List[int]]) -> int:
        rows=len(grid)
        cols=len(grid[0])
        count=0
        visited=set()
        def dfs(r,c):
            for ni,nj in [(0,-1),(0,1),(1,0),(-1,0)]:
                new_i=ni+r
                new_j=nj+c
                if 0<=new_i<rows and 0<=new_j<cols and grid[new_i][new_j]==0 and (new_i,new_j) not in visited:
                    visited.add((new_i,new_j))
                    dfs(new_i,new_j)
        for i in [0,rows-1]:
            for j in range(cols):
                if grid[i][j]==0 and (i,j) not in visited:
                    visited.add((i,j))
                    dfs(i,j)
        for i in [0,cols-1]:
            for j in range(rows):
                if grid[j][i]==0 and (j,i) not in visited:
                    visited.add((j,i))
                    dfs(j,i)
        for i in range(rows):
            for j in range(cols):
                if grid[i][j]==0 and (i,j) not in visited:
                    dfs(i,j)
                    count+=1
        return count

        rows = len(grid)
        cols = len(grid[0])
        count = 0
        visited = set()

        def dfs(r, c):
            if (r < 0 or c < 0 or r == rows or c == cols):
                return 0

            if grid[r][c] == 1 or (r, c) in visited:
                return 1  # This cell is water or has been visited

            visited.add((r, c))
            return min(
                dfs(r+1,c),
                dfs(r-1,c),dfs(r,c-1),
                dfs(r,c+1),
            )

        for r in range(rows):
            for c in range(cols):
                if grid[r][c] == 0 and (r, c) not in visited:
                    count+=dfs(r,c)

        return count

18100