<p>You are given an <code>m x n</code> binary matrix <code>grid</code>, where <code>0</code> represents a sea cell and <code>1</code> represents a land cell.</p>

<p>A <strong>move</strong> consists of walking from one land cell to another adjacent (<strong>4-directionally</strong>) land cell or walking off the boundary of the <code>grid</code>.</p>

<p>Return <em>the number of land cells in</em> <code>grid</code> <em>for which we cannot walk off the boundary of the grid in any number of <strong>moves</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/enclaves1.jpg" style="width: 333px; height: 333px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/enclaves2.jpg" style="width: 333px; height: 333px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All 1s are either on the boundary or can reach the boundary.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        def dfs(i, j, flag=False):

            if grid[i][j] == 0:
                return
            if flag:
                self.count += 1
            
            visited[i][j] = True
            neighbors = [(i, max(j-1, 0)), 
                         (i, min(j+1, n-1)), 
                         (max(i-1, 0), j), 
                         (min(i+1, m-1), j)]
            
            for x, y in neighbors:
                if not visited[x][y]:
                    dfs(x, y, flag)

        m = len(grid)
        n = len(grid[0])
        self.count = 0
        visited = [[False for _ in range(n)] for _ in range(m)]


        for i in range(m):
            if not visited[i][0]:
                dfs(i, 0)

        for i in range(m):
            if not visited[i][n-1]:
                dfs(i, n-1)

        for j in range(n):
            if not visited[0][j]:
                dfs(0, j)

        for j in range(n):
            if not visited[m-1][j]:
                dfs(m-1, j)

        # print(visited)
        for i in range(m):
            for j in range(n):
                if not visited[i][j]:
                    dfs(i, j, True)
        
        return self.count
56096