<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>

import numpy as np

class Solution:

    def __init__(self):
        self.island_area = 0

    def numEnclaves(self, grid: List[List[int]]) -> int:
        shape = np.shape(grid)

        # flip all islands on the boundary to water
        for i in range(shape[0]):
            for j in range(shape[1]):
                if (grid[i][j] == 1) and (i ==0 or i==shape[0]-1 or j==0 or j==shape[1]-1):
                    self.dfs(grid, shape, i, j)

        for i in range(shape[0]):
            for j in range(shape[1]):
                if grid[i][j] == 1:
                    self.island_area += 1

        return self.island_area
    

    def dfs(self, grid, shape, i, j):

        if (i<0) or (j<0) or (i >= shape[0]) or (j >= shape[1]):
            return

        if grid[i][j] == 0:
            return

        grid[i][j] = 0
        self.dfs(grid, shape, i-1, j)
        self.dfs(grid, shape, i+1, j)
        self.dfs(grid, shape, i, j-1)
        self.dfs(grid, shape, i, j+1)
39550