<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 key(self, x,y):
        return f'{x},{y}'

    def _dfs(self, ci,cj, grid,n,m,vis):
        vis.add(self.key(ci,cj))

        for nx,ny in [(ci-1,cj),(ci+1,cj),(ci,cj-1),(ci,cj+1)]:
            if nx not in range(n) or ny not in range(m):
                continue
            if grid[nx][ny]==1 and self.key(nx,ny) not in vis:
                self._dfs(nx,ny,grid,n,m,vis)


    def numEnclaves(self, grid: List[List[int]]) -> int:
        n,m = len(grid), len(grid[0])
        cnt=0
        for i in range(n):
            for j in range(m):
                if grid[i][j]==1:
                    cnt+=1

        vis=set()
        for i in range(n):
            if grid[i][0]==1 and self.key(i,0) not in vis:
                self._dfs(i,0,grid,n,m,vis)
            if grid[i][m-1]==1 and self.key(i,m-1) not in vis:
                self._dfs(i,m-1,grid,n,m,vis)

        for j in range(m):
            if grid[0][j]==1 and self.key(0,j) not in vis:
                self._dfs(0,j,grid,n,m,vis)
            if grid[n-1][j]==1 and self.key(n-1,j) not in vis:
                self._dfs(n-1,j,grid,n,m,vis)
        
        return cnt-len(vis)
        

        


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