<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:
        count=0
        def dfs(i,j):
            # print(i,j)
            visited[i][j]=1
            for k,l in ((i+1,j),(i,j+1),(i-1,j),(i,j-1)):
                if 0<=k<len(grid) and 0<=l<len(grid[0]) and grid[k][l]==1 and visited[k][l]==0:
                    dfs(k,l)
            return
        visited = [[0]*len(grid[0]) for _ in range(len(grid))]
        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(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(1,len(grid)-1):
            for j in range(1,len(grid[0])-1):
                if grid[i][j]==1 and visited[i][j]==0:
                    count+=1


        return count
        

        
        
        
        
        
50225