<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 = len(grid)
        n = len(grid[0])

        def boundary_check(i,j):
            if i == m -1 or j == n-1 or i==0 or j==0:
                return True
            else:
                return False

        def neighbour_list(i,j):
            ans = []
            if i+1 < m:
                ans.append((i+1,j))
            if i > 0:
                ans.append((i-1,j))
            if j +1 < n:
                ans.append((i,j+1))
            if j > 0:
                ans.append((i,j-1))
            return ans

        dfs_q = []

        i = 0
        j = 0

        curr_boundary = False
        curr_count = 0
        island_status = True
        ans = 0
        for i in range(m):
            for j in range(n):

                if grid[i][j] == -1 or grid[i][j] == 0:
                    continue
                
                curr_boundary = False
                curr_count = 0
                island_status = True

                dfs_q.append((i,j))
                grid[i][j] = -1

                while dfs_q:
                    r,c = dfs_q.pop()
                    #Mark visited
                    grid[r][c] = -1
                    if boundary_check(r,c):
                        island_status = False
                    #grid[i][j] = -1
                    curr_count  +=1
                    for rn,cn in neighbour_list(r,c):
                        #print("adding row  = ",rn)
                        #print("adding col  = ",cn)

                        #Add to queue if not visited
                        if grid[rn][cn] == 1:
                            dfs_q.append((rn,cn))
                            grid[rn][cn] = -1

                if island_status:
                    ans+=curr_count

        return ans    
                


        
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