<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, board: List[List[int]]) -> int:
        n = len(board)
        m = len(board[0])

        visited = [[0 for _ in range(m)] for _ in range(n)]

        def dfs(start):
            ci, cj = start
            visited[ci][cj] = 1
            
            delRow = [-1, 0, +1, 0]
            delCol = [0, +1, 0, -1]

            for x in range(4):
                ni = ci + delRow[x]
                nj = cj + delCol[x]

                if 0<= ni < n and 0 <= nj < m and visited[ni][nj] == 0 and board[ni][nj] == 1:
                    dfs((ni, nj))
        
        # Boundary traversal 
        # First and last row
        for i in range(m):
            if board[0][i] == 1 and visited[0][i] == 0:
                dfs((0, i))
            if board[n-1][i] == 1 and visited[n-1][i] == 0:
                dfs((n-1, i))
        # First and last column
        for i in range(n):
            if board[i][0] == 1 and visited[i][0] == 0:
                dfs((i, 0))
            
            if board[i][m-1] == 1 and visited[i][m-1] == 0:
                dfs((i, m-1))

        # Now the visited array will be updated with 1's wherever thhere are 0's connected to the Boundary
        # Leaving them, we can count all other 1's for not able to reach the boundary
        count = 0

        for i in range(n):
            for j in range(m):
                if board[i][j] == 1 and visited[i][j] == 0:
                    count += 1
        return count




        



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