<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, mat: List[List[int]]) -> int:
        def dfs(row,col):
            vis[row][col]=1
            lst1=[-1, 0, 1, 0]
            lst2=[0, 1, 0, -1]
            for i in range(4):
                a,b=row+lst1[i],col+lst2[i]
                if a>=0 and a<n and b>=0 and b<m and vis[a][b]==0 and mat[a][b]==1:
                    dfs(a,b)
        n=len(mat)
        m=len(mat[0])
        vis=[[0 for _ in range(m)] for _ in range(n)]
        for i in range(n):
            for j in range(m):
                if i==0 or j==0 or i==n-1 or j==m-1:
                    if(mat[i][j]==1):
                        dfs(i,j)
        c=0            
        for i in range(len(mat)):
            for j in range(len(mat[0])):
                if vis[i][j]!=mat[i][j]:
                    c+=1   
        return c                    


        
44888