<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 dfs(self, grid, r, c):
        grid[r][c] = 0

        dy = [0, 1, 0, -1]
        dx = [1, 0, -1, 0]

        ans = 0
        for i in range(len(dy)):
            ny, nx = r + dy[i], c + dx[i]
            if 0 <= ny < len(grid) and 0 <= nx < len(grid[0]) and grid[ny][nx]:
                ans += self.dfs(grid, ny, nx)

        return ans + 1

    def numEnclaves(self, grid: List[List[int]]) -> int:
        total_lands = sum(1 for r in range(len(grid)) for c in range(len(grid[0])) if grid[r][c] == 1)

        row, col = len(grid), len(grid[0])
        top = sum(self.dfs(grid, 0, c) for c in range(col) if grid[0][c] == 1)
        left = sum(self.dfs(grid, r, 0) for r in range(row) if grid[r][0] == 1)
        right = sum(self.dfs(grid, r, col-1) for r in range(row) if grid[r][col-1] == 1)
        bottom = sum(self.dfs(grid, row-1, c) for c in range(col) if grid[row-1][c] == 1)

        return (total_lands - top - left - right - bottom)
43286