<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 __init__(self):
        self.visited = []
    def dfs(self,x,y,grid):
        self.visited[x][y] = 1
        delR = [-1,0,1,0]
        delC = [0,1,0,-1]
        for i in range(4):
            nrow = x + delR[i]
            ncol = y + delC[i]
            if 0 <= nrow < len(grid) and 0 <= ncol < len(grid[0]) and not self.visited[nrow][ncol] and grid[nrow][ncol]:
                self.dfs(nrow,ncol,grid)
    def numEnclaves(self, grid: List[List[str]]) -> None:
        r,c = len(grid),len(grid[0])
        self.visited = [[0]*c for _ in range(r)]
        for i in range(r):
            for j in range(c):
                if (i == 0 or i == r-1 or j == 0 or j == c-1) and grid[i][j]:
                    self.dfs(i,j,grid)
        print(self.visited)
        cnt = 0
        for i in range(r):
            for j in range(c):
                if grid[i][j] and not self.visited[i][j]:
                    cnt+=1
        return cnt
                
        
        
42753