<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 bfs(self, cell, result, grid):
        # r,c = cell
        m,n = len(grid), len(grid[0])
        q = deque()
        q.append(cell)

        while q:
            r,c = q.pop()
            for i, j in [[r-1,c], [r+1,c], [r, c-1], [r, c+1]]:
                if i<0 or j<0 or i>=m or j>=n or grid[i][j]==0 or (i,j) in result:
                    continue
                else:
                    result[(i,j)] = 1
                    q.append((i,j))
    
    def numEnclaves(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        ones = 0
        boundary = []
        for i in range(m):
            for j in range(n):
                if grid[i][j]==1:
                    ones+=1
                    if i==0 or i==m-1:
                        boundary.append((i,j))
        if ones==0:
            return 0
        
        for i in range(m):
            if grid[i][0]==1:
                boundary.append((i,0))
            if grid[i][n-1]==1:
                boundary.append((i,n-1))

        if not boundary:
            return ones
        result = {}

        for cell in boundary:
            if cell not in result:
                result[cell] = 1
                self.bfs(cell, result, grid)
        
        # print(boundary)
        return ones-len(result)
26206