<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, grid: List[List[int]]) -> int:
        m ,n  = len(grid) , len(grid[0])
        stack = []
        visited = set()
        dirs = [[1,0],[-1,0],[0,1],[0,-1]]
        def dfs():
            nonlocal m,n
            while stack:
                row ,col = stack.pop()
                for d in dirs:
                    r,c = row+d[0] , col +d[1]
                    if r<m and r>=0 and c<n and c>=0 and grid[r][c] ==1 and (r,c) not in visited:
                        visited.add((r,c))
                        stack.append([r,c])

        for j in range(n):
            if grid[0][j] ==1 and (0,j) not in visited:
                print(" boundary cell ", 0,j)
                visited.add((0,j))
                stack.append([0,j])
                dfs()
            if grid[m-1][j] ==1 and (m-1,j) not in visited:
                print(" boundary cell ", m-1,j)
                visited.add((m-1, j))
                stack.append([m-1, j])
                dfs()
        for i in range(m):
            if grid[i][0] ==1 and (i,0) not in visited:
                print(" boundary cell ", i,0)
                visited.add((i,0))
                stack.append([i,0])
                dfs()
            if grid[i][n-1] ==1 and (i,n-1) not in visited:
                print(" boundary cell ", i,n-1)
                visited.add((i,n-1))
                stack.append([i,n-1])
                dfs()

        enclaves =0
        for i in range(m):
            for j in range(n):
                if grid[i][j] ==1 and (i,j) not in visited:
                    enclaves+=1
                    print("unvisited 1s at ", i,j)
                    visited.add((i,j))
        return enclaves

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