<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>

import numpy as np
class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        # res=0
        # def dfs(r,c):
        #     for (i,j) in((r+1,c),(r-1,c),(r,c+1),(r,c-1)):
        #         if 0<=i<len(grid) and 0<=j<len(grid[0]) and grid[i][j]==1:
        #             grid[i][j]=0
        #             dfs(i,j)
        # for i in range(len(grid)):
        #     if grid[i][0]==1:
        #         dfs(i,0)
        #     if grid[i][-1]==1:
        #         dfs(i,len(grid[0])-1)
        # for i in range(len(grid[0])):
        #     if grid[0][i]==1:
        #         dfs(0,i)
        #     if grid[-1][i]==1:
        #         dfs(len(grid)-1,i)
        # for i in range(1,len(grid)-1):
        #     for j in range(1,len(grid[0])-1):
        #         if grid[i][j]==1:
        #             res+=1
        # return res
        res=0
        m=len(grid)
        n=len(grid[0])
        visited=np.zeros((m,n))
        def dfs(r,c):
            if r==0 or r==m-1 or c==0 or c==n-1:
                return 0
            area=1
            left,right,up,down=-1,-1,-1,-1
            if grid[r][c+1]==1 and visited[r][c+1]==0:
                visited[r][c+1]=1
                left=dfs(r,c+1)
            if grid[r][c-1]==1 and visited[r][c-1]==0:
                visited[r][c-1]=1
                right=dfs(r,c-1)
            if grid[r+1][c]==1 and visited[r+1][c]==0:
                visited[r+1][c]=1
                down=dfs(r+1,c)
            if grid[r-1][c]==1 and visited[r-1][c]==0:
                visited[r-1][c]=1
                up=dfs(r-1,c)
            if up==0 or down==0 or left==0 or right==0:
                return 0
            return max(left,0)+max(right,0)+max(up,0)+max(down,0)+1
            # for (i,j) in ((r+1,c),(r-1,c),(r,c+1),(r,c-1)):
            #     if grid[i][j]==1 and visited[i][j]==0:
            #         visited[i][j]=1
            #         val=dfs(i,j)
            #         if val!=0:
            #             area+=val
            #         else:
            #             return 0
            return area
        res=0
        for i in range(1,m-1):
            for j in range(1,n-1):
                if grid[i][j]==1 and visited[i][j]==0:
                    visited[i][j]=1
                    res+=dfs(i,j)
        return res
                    
        
43820