<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 bfs(self, x, y, visited, grid):
        visited[x][y] = 1
        m, n = len(grid), len(grid[0])
        q = deque()
        q.append([x, y])
        
        flag = True
        count = 1
        
        while q:
            node = q.popleft()
            x, y = node[0], node[1]

            for i, j in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
                nx = x + i
                ny = y + j
                if nx >= 0 and ny >= 0 and nx < m and ny < n:
                    if visited[nx][ny] == 0 and grid[nx][ny] == 1:
                        visited[nx][ny] = 1
                        count += 1
                        q.append([nx, ny])
                else:
                    flag = False
        
        return [flag, count]

    def dfs(self, x, y, visited, grid, count):
        visited[x][y] = 1
        count += 1
        # print("Count here:", count)
        m, n = len(grid), len(grid[0])

        for i, j in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
            nx = x + i
            ny = y + j
            if nx >= 0 and ny >= 0 and nx < m and ny < n:
                if visited[nx][ny] == 0 and grid[nx][ny] == 1:
                    count = self.dfs(nx, ny, visited, grid, count)
            else:
                count = float("-inf")
        # print("Here?", count)
        return count

    def numEnclaves(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        visited = [[] for i in range(m)]
        for i in visited:
            for j in range(n):
                i.append(0)
        # print(visited)

        ans = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if visited[i][j] == 0 and grid[i][j] == 1:
                    print("i:", i, "j:", j)
                    # res = self.bfs(i, j, visited, grid, count)
                    # print(res)
                    count = 0
                    count = self.dfs(i, j, visited, grid, count)
                    if count > 0:
                        ans += count
                    # if res[0]:
                    #     ans += res[1] 
        
        return ans
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