<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>

def dfs(r, c, board, vis):
    vis.add((r, c))
    dx = [-1, 1, 0, 0]
    dy = [0, 0, -1, 1]
    
    for x in range(4):
        nr = r + dx[x]
        nc = c + dy[x]
        if nr >= 0 and nc >= 0 and nr < len(board) and nc < len(board[0]) and board[nr][nc] == 1 and (nr, nc) not in vis:
            dfs(nr, nc, board, vis)

class Solution:
    def numEnclaves(self, board: List[List[int]]) -> int:
        m = len(board)
        n = len(board[0])
        vis = set()

        for i in range(m):
            if board[i][0] == 1:
                dfs(i, 0, board, vis)
            if board[i][n - 1] == 1:
                dfs(i, n - 1, board, vis)
        
        for j in range(n):
            if board[0][j] == 1:
                dfs(0, j, board, vis)
            if board[m - 1][j] == 1:
                dfs(m - 1, j, board, vis)
        
        ans = 0
        for i in range(m):
            for j in range(n):
                if board[i][j] == 1 and(i, j) not in vis:
                    ans += 1
        
        return ans


        
49158