<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])
        v = [[False] * m for _ in range(n)]
        
        def dfs(row, col, c, isConnectedBoundary):
            v[row][col] = True
            c[0] += 1
            if row == 0 or col == 0 or row == n - 1 or col == m - 1:
                isConnectedBoundary[0] = True
            delrow = [-1, 0, 1, 0]
            delcol = [0, 1, 0, -1]
            
            for i in range(4):
                nrow, ncol = row + delrow[i], col + delcol[i]
                
                if 0 <= nrow < n and 0 <= ncol < m and not v[nrow][ncol] and grid[nrow][ncol] == 1:
                    dfs(nrow, ncol, c, isConnectedBoundary)
        
        ans = 0
        for i in range(n):
            for j in range(m):
                if not v[i][j] and grid[i][j] == 1:
                    c = [0]  
                    isConnectedBoundary = [False] 
                    dfs(i, j, c, isConnectedBoundary)
                    if not isConnectedBoundary[0]: 
                        ans += c[0]  
        
        return ans
  

                    
                
                    
            
        
        
46489