<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>

from typing import List

class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        n = len(grid)
        m = len(grid[0])
        rank = [1 for i in range(n * m + 1)]
        parent = [i for i in range(n * m + 1)]
        
        def find(p):
            if parent[p] != p:
                parent[p] = find(parent[p])
            return parent[p]

        def union(p, q):
            rootP = find(p)
            rootQ = find(q)
            if rootP == rootQ:
                return
            if rank[rootP] > rank[rootQ]:
                parent[rootQ] = rootP
            elif rank[rootP] < rank[rootQ]:
                parent[rootP] = rootQ
            else:
                parent[rootQ] = rootP
                rank[rootP] += 1

        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        virtual_node = n * m  # Extra node representing all boundary connections

        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1:
                    for dx, dy in directions:
                        x = i + dx
                        y = j + dy
                        if 0 <= x < n and 0 <= y < m and grid[x][y] == 1:
                            union(i * m + j, x * m + y)
                    if i == 0 or i == n - 1 or j == 0 or j == m - 1:
                        union(i * m + j, virtual_node)

        # Count the number of enclaves
        enclave_count = 0
        for i in range(n):
            for j in range(m):
                if grid[i][j] == 1 and find(i * m + j) != find(virtual_node):
                    enclave_count += 1

        return enclave_count

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