<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:

     

        def dfs(row: int, col: int , visited):
            rows = [ -1 , 1 , 0 , 0]
            cols = [ 0 , 0 , -1 , 1]
            

            for i in range(4) :

                newr = row + rows[i]
                newc = col + cols[i]

                if 0 <= newr < m and 0 <= newc < n and visited[newr][newc] == 0 and grid[newr][newc] == 1 :

                    visited[newr][newc] = 1
                    dfs(newr , newc , visited)


        m = len(grid)
        n = len(grid[0])

        visited = [ [0 for i in range(n)] for j in range(m)]

        for i in range(m) :

            for j in range(n) :

                if i == 0 or i == m - 1 :

                    if grid[i][j] == 1 and visited[i][j] == 0 :

                        visited[i][j] = 1
                        dfs( i , j , visited )
                else :

                    if grid[i][0] == 1 and visited[i][0] == 0 :

                        visited[i][0] = 1
                        dfs( i , 0 , visited )
                    
                    elif grid[i][n-1] == 1 and visited[i][n-1] == 0 :
                        
                        visited[i][n-1] = 1
                        dfs( i , n-1 , visited)
                    
                    
                    
        
        c = 0


        for i in range(1,m) :

            for j in range(1,n) :

                if grid[i][j] == 1 and visited[i][j] == 0 :
                    c += 1
        return c



                    



        
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