<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:
        n, m = len(grid), len(grid[0])
        vis = [[0]*m for _ in range(n)]
        drow = [-1,0,+1,0]
        dcol = [0,+1,0,-1]
        for i in range(n):
            if(not(vis[i][0]) and grid[i][0]==1):
                self.dfs(i,0,vis,grid,n,m,drow,dcol)
            if(not(vis[i][m-1]) and grid[i][m-1]==1):
                self.dfs(i,m-1,vis,grid,n,m,drow,dcol)    
        for j in range(m):
            if(not(vis[0][j]) and grid[0][j]):
                self.dfs(0,j,vis,grid,n,m,drow,dcol)
            if(not(vis[n-1][j]) and grid[n-1][j]):
                self.dfs(n-1,j,vis,grid,n,m,drow,dcol)                   
        cnt = 0
        for i in range(n):
            for j in range(m):
                if(not(vis[i][j]) and grid[i][j]==1):
                    cnt+=1
        return cnt            

    def dfs(self,row,col,vis,grid,n,m,drow,dcol):
        vis[row][col] = 1
        for i in range(4):
            nrow, ncol = row+drow[i], col+dcol[i]
            if(-1<nrow<n and -1<ncol<m and not(vis[nrow][ncol]) and grid[nrow][ncol]==1 ):
                self.dfs(nrow,ncol,vis,grid,n,m,drow,dcol)


        
35814