<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.visited = None
    def get_neighbours(self, node, board):
        res = []
        x, y = node
        for dx, dy in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
            xx, yy = x + dx, y + dy
            if xx < 0 or yy < 0 or xx >= len(board) or yy >= len(board[0]) or board[xx][yy] == 0:
                continue
            res.append((xx, yy))
        return res
    def dfs(self, node, neighbours, board):
        self.visited.add(node)
        for neighbour in neighbours:
            if neighbour not in self.visited:
                self.dfs(neighbour, self.get_neighbours(neighbour, board), board)
        return
    def numEnclaves(self, board: List[List[int]]) -> int:
        self.visited = set()

        for i in [0, len(board)-1]:
            for j in range(len(board[0])):
                if board[i][j] == 1:
                    self.dfs((i, j), self.get_neighbours((i, j), board), board)
        
        for i in range(len(board)):
            for j in [0, len(board[0])-1]:
                if board[i][j] == 1:
                    self.dfs((i, j), self.get_neighbours((i, j), board), board)  
        ans = 0
        for i in range(len(board)):
            for j in range(len(board[0])):
                if (i, j) not in self.visited and board[i][j] == 1:
                    ans += 1
        return ans   
53428