<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:
        width = len(grid[0])
        height = len(grid)
        visited = set() 
        count = 0

        for i in range(height):
            for j in range(width):
                if(grid[i][j] == 1):
                    count += 1

        def dfs(row, col):
            if((row, col) in visited or row < 0 or row not in range(height) or col < 0 or col not in range(width) or grid[row][col] == 0):
                return
            visited.add((row, col))

            directions = [[0, 1], [1, 0], [0, -1], [-1, 0]]
            dfs(row + directions[0][0], col + directions[0][1])
            dfs(row + directions[1][0], col + directions[1][1])
            dfs(row + directions[2][0], col + directions[2][1])
            dfs(row + directions[3][0], col + directions[3][1])


        for i in range(height):
            if(grid[i][0] == 1):
                dfs(i, 0) 
            if(grid[i][width-1] == 1):
                dfs(i, width-1)

        for j in range(width):
            if(grid[0][j] ==1):
                dfs(0, j) 
            if(grid[height-1][j] == 1):
                dfs(height - 1, j)

        return(count - len(visited))
59833