<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, g: List[List[int]]) -> int:
        row=len(g)
        col=len(g[0])
        p=[(-1,0),(1,0),(0,1),(0,-1)]
        visited=[[0]*col for i in range(row)]
        count=[0]
        def dfs(i,j):
            s=[(i,j)]
            while s:
                r,c=s.pop()
                visited[r][c]=1
                count[0]+=1
                for i in p:
                    if r+i[0]>=0 and r+i[0]<row and c+i[1]>=0 and c+i[1]<col and g[r+i[0]][c+i[1]]==1 and visited[r+i[0]][c+i[1]]==0:
                        s.append((r+i[0],c+i[1]))
                        visited[r+i[0]][c+i[1]]=1
        for k in range(col):
            if g[0][k]==1 and visited[0][k]==0:
                dfs(0,k)
        for k in range(col):
            if g[row-1][k]==1 and visited[row-1][k]==0:
                dfs(row-1,k)
        for k in range(1,row-1):
            if g[k][0]==1 and visited[k][0]==0:
                dfs(k,0)
        for k in range(1,row-1):
            if g[k][col-1]==1 and visited[k][col-1]==0:
                dfs(k,col-1)
        ones=0
        for i in g:
            for j in i:
                if j==1:
                    ones+=1
        return ones-count[0]
19801