<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:
        visited=[[0 for j in range(len(grid[0]))] for i in range(len(grid))]
        def dfs(node):
            r,c=node
            visited[r][c]=1
            neigh=[(r-1,c),(r,c+1),(r+1,c),(r,c-1)]
            for nr,nc in neigh:
                if(nr in range(len(grid)) and nc in range(len(grid[0])) and grid[nr][nc]==1 and visited[nr][nc]==0):
                    dfs((nr,nc))
        for j in range(len(grid[0])):
            if(visited[0][j]==0 and grid[0][j]==1):
                dfs((0,j))
            if(visited[len(grid)-1][j]==0 and grid[len(grid)-1][j]==1):
                dfs((len(grid)-1,j))
        for i in range(len(grid)):
            if(visited[i][0]==0 and grid[i][0]==1):
                dfs((i,0))
            if(visited[i][len(grid[0])-1]==0 and grid[i][len(grid[0])-1]==1):
                dfs((i,len(grid[0])-1))
        cnt=0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if(grid[i][j]==1 and visited[i][j] == 0):
                    cnt+=1
        return cnt
57698