<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 dfs(self, row, col, isVisited, grid):
        if isVisited[row][col] or grid[row][col] != 1: return
        isVisited[row][col] = True
        dirArr = [[0,1],[0,-1],[1,0],[-1,0]]
        for i, j in dirArr:
            r, c = row+i, col+j
            if r < 0 or c < 0 or r >= len(grid) or c >= len(grid[0]):
                continue
            self.dfs(r, c, isVisited, grid)


    def numEnclaves(self, grid: List[List[int]]) -> int:
        rowSize, colSize = len(grid), len(grid[0])
        isVisited = [ [False] * colSize  for _ in range(rowSize) ]
        for col in range(colSize):
            self.dfs(0, col, isVisited, grid)
            self.dfs(rowSize-1, col, isVisited, grid)
        for row in range(rowSize):
            self.dfs(row, 0, isVisited, grid)
            self.dfs(row, colSize-1, isVisited, grid)
        cnt = 0
        for r in range(rowSize):
            for c in range(colSize):
                if grid[r][c] == 1 and not isVisited[r][c]:
                    cnt += 1
        return cnt
        
            
        
56630