<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 __init__(self):
        self.directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]
        self.grid = [[]]
        self.visited = set()
        self.m = 0
        self.n = 0
    
    def dfs(self, r, c):
        self.visited.add((r, c))
        ans = 1
        for dx, dy in self.directions:
            nr, nc = r + dx, c + dy
            if (0 <= nr < self.m 
                and 0 <= nc < self.n 
                and (nr, nc) not in self.visited 
                and self.grid[nr][nc] == 1):
                ans += self.dfs(nr, nc)
        return ans

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

        boundary_land = 0
        total_land = 0
        for i in range(self.m):
            for j in range(self.n):
                total_land += grid[i][j]
                if (grid[i][j] == 1 
                    and (i in [0, self.m - 1] or j in [0, self.n - 1])
                    and (i, j) not in self.visited):
                    boundary_land += self.dfs(i, j)
                    
        return total_land - boundary_land

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