<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:
        
        border = set()

        for i, row in enumerate(grid):
            for j, val in enumerate(row):
                on_boarder = i == 0 or i == len(grid) - 1 or j == 0 or j == len(row) - 1

                if on_boarder and val == 1 and (i,j) not in border:
                    self.dfs(grid, (i,j), border)

        count = 0
        for i, row in enumerate(grid):
            for j, val in enumerate(row):

                if val == 1 and (i,j) not in border:
                    count += 1
        
        return count


    def dfs(self, graph, start, border):
        queue = [start]
        visited = set()

        while queue:
            node = queue.pop()
            if node in visited: continue
            visited.add(node)
            border.add(node)

            i, j = node

            # up
            if i > 0 and graph[i-1][j] == 1:
                queue.append((i-1, j))
            # down
            if i < len(graph) - 1 and graph[i+1][j] == 1:
                queue.append((i+1, j))
            # left
            if j > 0 and graph[i][j-1] == 1:
                queue.append((i, j-1))
            # right
            if j < len(graph[i]) - 1 and graph[i][j+1] == 1:
                queue.append((i, j+1)) 
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