<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 __init__(self):
        self.visited = set()
        self.directions = [[0, -1], [0, 1], [-1, 0], [1, 0]]
        self.ans = 0
        self.total = 0
        
    def numEnclaves(self, grid: List[List[int]]) -> int:
        numRows = len(grid)
        numCols = len(grid[0])
        
        for i in range(numRows):
            for j in range(numCols):
                if grid[i][j] == 1 and (i, j) not in self.visited:
                    self.total = 0
                    off = self.bfs(i, j, grid)
                    if not off:
                        self.ans += self.total
        return self.ans
                        
    def bfs(self, i, j, grid):
        numRows = len(grid)
        numCols = len(grid[0])
        deq = collections.deque()
        deq.append((i, j))
        self.visited.add((i, j))
        off = False
        
        while deq:
            row, col = deq.popleft()
            
            if row == 0 or row == numRows - 1 or col == 0 or col == numCols - 1:
                off = True
            
            self.total += 1
            
            for dr, dc in self.directions:
                newRow = row + dr
                newCol = col + dc
                
                if newRow < 0 or newCol < 0 or newCol >= numCols or newRow >= numRows or grid[newRow][newCol] == 0:
                        continue
                
                if (newRow, newCol) not in self.visited:
                    deq.append((newRow, newCol))
                    self.visited.add((newRow, newCol))
        return off   
    
                
            
        
        
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