<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 neighbour(cur):
            li=[]
            deltas=[(0,1),(0,-1),(1,0),(-1,0)]
            for d in deltas:
                newx,newy=(cur[0]+d[0],cur[1]+d[1])
                if newx>=0 and newx<m and newy>=0 and newy<n and\
                board[newx][newy]==1:
                    li.append((newx,newy))
            ##print(li)
            return li
        m=len(board)
        n=len(board[0])
        q=deque()
        visited=set()
        for i in range(m):
            
                if board[i][0]==1:
                    q.append((i,0))
                    visited.add((i,0))
                if board[i][n-1]==1:
                    q.append((i,n-1))
                    visited.add((i,n-1))
        for j in range(n):
            if board[0][j]==1:
                q.append((0,j))
                visited.add((0,j))
            if board[m-1][j]==1:
                q.append((m-1,j))
                visited.add((m-1,j))
        #for each one call a dfs 
        
        
            
            
        
        
        while len(q)!=0:
            cur=q.popleft()
            
            
            print(f'cur {cur}')
            
            
            #board[cur[0]][cur[1]]='#'
            #visited.add(cur)
            for nei in neighbour(cur):
                
                    if nei not in visited:
                        print(f'nei  {nei}')
                        visited.add(nei)
                        q.append(nei)
                    #board[nei[0]][nei[1]]='#'
            #print(visited,empty)
        #print(q)
        #multi_bfs(q,board)
        #print(board)
        count=0
        for i in range(m):
            for j in range(n):
                if board[i][j]==1:
                    count+=1
        print(visited)
        print(count,len(visited),m,n)
        return count-len(visited)
25139