<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:
        def dfs(row, col):
            print(row, col)
            visited[row][col] = 1
            possible = {(0,1), (0, -1), (1, 0), (-1, 0)}
            for row_off_set, col_off_set in possible:
                new_row = row + row_off_set
                new_col = col + col_off_set
                if new_row >= 0 and new_row < len(board) and new_col >= 0 and new_col < len(board[0]) and visited[new_row][new_col] == 0 and board[new_row][new_col] == 1:
                    dfs(new_row, new_col)

        visited = [[0 for i in range(len(board[0]))] for j in range(len(board))] 
        for i in range(len(board[0])):
            if (visited[0][i] == 0 and board[0][i] == 1):
                dfs(0, i)
            
            if (visited[len(board) - 1][i] == 0 and board[len(board) - 1][i] == 1):
                dfs(len(board) - 1, i)
        

        for j in range(len(board)):
            if (visited[j][0] == 0 and board[j][0] == 1):
                dfs(j, 0)
            
            if (visited[j][len(board[0]) - 1] == 0 and board[j][len(board[0]) - 1] == 1):
                dfs(j, len(board[0]) - 1)
        
        count = 0
        for i in range(len(board)):
            for j in range(len(board[0])):
                if board[i][j] == 1 and visited[i][j] == 0:
                    count+= 1
        return count


        #TC: O(N * M) + O(N) + O(M) worst case if all are O
        #SC: O(N * M)        

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