<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:
        if not grid:
            return 0
        
        ROWS, COLS = len(grid), len(grid[0])
        visit = set()

        #dfs
        def dfs(r,c):
            if (min(r,c) < 0 or r == ROWS or c == COLS or grid[r][c] == 0 or (r,c) in visit):
                return 0
            
            visit.add((r,c))
            res = 1
            # direct = [[0,1],[0,-1],[1,0],[-1,0]]
            # for dr, dc in direct:
            #     res += dfs(r + dr, c + dc)
            res += dfs(r+1,c)
            res += dfs(r-1,c)
            res += dfs(r,c-1)
            res += dfs(r,c+1)
            return res

        #bfs
        def bfs(r, c):
            queue = deque([(r, c)])
            visit.add((r, c))
            res = 1
            
            while queue:
                r, c = queue.popleft()
                direct = [[0,1],[0,-1],[1,0],[-1,0]]
                for dr, dc in direct:
                    nr, nc = r + dr, c + dc
                    if (nr in range(ROWS) and nc in range(COLS) and grid[nr][nc] == 1 and (nr,nc) not in visit):
                        visit.add((nr, nc))
                        queue.append((nr, nc))
                        res += 1
            
            return res

        land, borderland = 0, 0
        for r in range(ROWS):
            for c in range(COLS):
                land += grid[r][c] == 1
                if (grid[r][c] == 1 and (r,c) not in visit and (c in [0, COLS - 1] or r in [0, ROWS - 1])): # dfs around border that has 1
                    # borderland += dfs(r,c)
                    borderland += bfs(r,c)
        
        return land - borderland
        
25673