<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:
        
        m,n = len(grid), len(grid[0])
        count = 0
        for i in [0, m-1]:
            for j in range(n):
                if grid[i][j] == 0:
                    self.ff(grid, m,n,i,j)
        for i in range(m):
            for j in [0,n-1]:
                if grid[i][j] == 0:
                    self.ff(grid, m, n, i, j)
        
        for i in range(1, m-1):
            for j in range(1, n-1):
                if grid[i][j] == 0:
                    count += 1
                    self.ff(grid, m,n, i, j)
        
        return count
        
        
    def ff(self, grid, m, n, i, j):
        if grid[i][j] == 1:
            return
        grid[i][j] = 1
        
        for i1,j1 in self.nbr(m,n,i,j):
            if grid[i1][j1] == 0:
                self.ff(grid, m,n,i1,j1)
                
            
        return
    def nbr(self, m,n,i,j):
        nb = []
        
        if i > 0: nb.append((i-1, j))
        if i < m-1: nb.append((i+1, j))
        if j > 0: nb.append((i, j-1))
        if j < n-1: nb.append((i, j+1))
            
        return nb
        
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