<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:
        def dfs(node,grid,vis):
            dR = [-1,0,1,0]
            dC = [0,1,0,-1]
            vis.add((node[0],node[1]))
            grid[node[0]][node[1]] = 0
            for i in range(4):
                nr = node[0] + dR[i]
                nc = node[1] + dC[i]
                if nr >= 0 and nc >= 0 and nr < len(grid) and nc < len(grid[0]) and (nr,nc) not in vis and grid[nr][nc] == 1:
                    dfs([nr,nc],grid,vis)
            
            
        vis = set()
        for i in range(len(grid)):
            if grid[i][0] == 1:
                dfs([i,0],grid,vis)
            if grid[i][len(grid[0])-1] == 1:
                dfs([i,len(grid[0])-1],grid,vis)
        for j in range(len(grid[0])):
            if grid[0][j] == 1:
                dfs([0,j],grid,vis)
            if grid[len(grid)-1][j] == 1:
                dfs([len(grid)-1,j],grid,vis)
        
        
        cnt = 0
        for i in grid:
            for j in i:
                if j == 1:
                    cnt+=1
        return cnt
        
        
53428