<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:
#         n = len(grid)
#         m = len(grid[0])
#         count = 0
#         def dfs(r,c):
#             grid[r][c] = 0
#             directions = [[1,0],[0,1],[-1,0],[0,-1]]
#             for dr,dc in directions:
#                 R = r+dr
#                 C = c+dc
#                 if R<0 or C<0 or R>=n or C>=m or grid[R][C]!=1:
#                     continue                
#                 dfs(R,C)
#         for i in range(n):
#             for j in range(m):
#                 if (i==0 or i==n-1 or j==0 or j==m-1) and grid[i][j]==1:
#                     dfs(i,j)
#         for i in range(n):
#             for j in range(m):
#                 if grid[i][j] == 1:
#                     count+=1
#         return count

class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])
        count = 0
        visited = [[0 for i in range(m)]for j in range(n)]
        def dfs(r,c,visited):
            grid[r][c] = 0
            visited[r][c] = 1
            directions = [[1,0],[0,1],[-1,0],[0,-1]]
            for dr,dc in directions:
                R = r+dr
                C = c+dc
                if R<0 or C<0 or R>=n or C>=m or grid[R][C] in visited or grid[R][C]!=1:
                    continue                
                dfs(R,C,visited)
        for i in range(n):
            for j in range(m):
                if (i==0 or i==n-1 or j==0 or j==m-1) and grid[i][j]==1:
                    dfs(i,j,visited)
        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1:
                    count+=1
        return count
58765