<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:
        m=len(grid)
        n=len(grid[0])
        vis=[[0 for i in range(n)]for j in range(m)]
        dir=[(0,1),(0,-1),(1,0),(-1,0)]
        def dfs(r,c):
            vis[r][c]=1
            for d in dir:
                nr=r+d[0]
                nc=c+d[1]
                if(nr>=0 and nr<m and nc>=0 and nc<n and grid[nr][nc]==1 and vis[nr][nc]==0):
                    dfs(nr,nc)


        for i in range(m):
            for j in range(n):
                if((i==0 or i==m-1 or j==0 or j==n-1) and grid[i][j]==1 and vis[i][j]==0):
                    print("i",i,"j",j)
                    dfs(i,j)
        print(vis)
        count=0
        for i in range(m):
            for j in range(n):
                if(grid[i][j]==1 and vis[i][j]==0):
                    count+=1

        return count
        

33145