<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, board: List[List[int]]) -> int:
        def dfs(x,y,alist):
            if 0<=x<=len(alist)-1 and 0<=y<=len(alist[0])-1 and alist[x][y] == 1:
                alist[x][y] = 0
                for (dx,dy) in [(-1,0),(1,0),(0,-1),(0,1)]:
                    if 0<=x+dx<=len(alist)-1 and 0<=y+dy<=len(alist[0])-1:
                        dfs(x+dx,y+dy,alist)
            

        for i in range(len(board)):
            if board[i][0] == 1:
                dfs(i,0,board)

            if board[i][len(board[0])-1] == 1:
                dfs(i,len(board[0])-1,board)

        for j in range(len(board[0])):
            if board[0][j] == 1:
                dfs(0,j,board)

            if board[len(board)-1][j] == 1:
                dfs(len(board)-1,j,board)
        cnt = 0
        for i in range(len(board)):
            for j in range(len(board[0])):
                if board[i][j] == 1:
                    cnt += 1
        return cnt
30476