<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 closedIsland2(self, grid: List[List[int]]) -> int:
        # IMPORTANT NOTE: the easiest way to deal with the boundary island problem is to first do DFS from the boundary, then inside!!!
        m, n = len(grid), len(grid[0])
        
        def dfs(grid, i, j):
            if i < 0 or i > m - 1 or j < 0 or j > n - 1 or grid[i][j] == 1:
                return
            grid[i][j] = 1  # mark as visited
            for (di, dj) in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                ni, nj = i + di, j + dj
                dfs(grid, ni, nj)

        for i in [0, m - 1]:
            for j in range(n):
                dfs(grid, i, j)
        for i in range(m):
            for j in [0, n - 1]:
                dfs(grid, i, j)
        count = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 0:
                    count += 1
                    dfs(grid, i, j)
        return count

    def closedIsland(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])

        def dfs(grid, i, j):
            if i < 0 or i > m - 1 or j < 0 or j > n - 1:
                return False
            if grid[i][j] == 1:
                return True
            grid[i][j] = 1  # mark as visited
            out = True
            for (di, dj) in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                ni, nj = i + di, j + dj
                out = dfs(grid, ni, nj) and out
            return out

        count = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 0:
                    if dfs(grid, i, j):
                        count += 1
        return count

17100