<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,i,j,grid,vis,n,m):
        row = [1,0,-1,0]
        col = [0,1,0,-1]
        for x,y in zip(row,col):
            if i+x>=0 and i+x<n and j+y>=0 and j+y<m and vis[i+x][j+y]==0 and grid[i+x][j+y]==1:
                vis[i+x][j+y] =1
                self.dfs(i+x,j+y,grid,vis,n,m)
    def numEnclaves(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])
        visited = [[0 for y in range(m)]for x in range(n)]
        for i in range(n):
            if grid[i][0] == 1 and visited[i][0] == 0:
                visited[i][0] = 1
                self.dfs(i,0,grid,visited,n,m)
            if grid[i][m-1] == 1 and visited[i][m-1] == 0:
                visited[i][m-1] = 1
                self.dfs(i,m-1,grid,visited,n,m)
        for j in range(m):
            if grid[0][j] == 1 and visited[0][j] == 0:
                visited[0][j] = 1
                self.dfs(0,j,grid,visited,n,m)
            if grid[n-1][j] == 1 and visited[n-1][j] == 0:
                visited[n-1][j] = 1
                self.dfs(n-1,j,grid,visited,n,m)
        ans = 0
        for i in range(1,n):
            for j in range(1,m):
                if visited[i][j] == 0 and grid[i][j]==1:
                    ans+=1
        return ans


        
59299