<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 UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))
        self.rank = [1]*n
        self.isConnectedToBoundary = [False]*n

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        
        if rootX != rootY:
            if self.rank[rootX] > self.rank[rootY]:
                self.parent[rootY] = rootX
                self.isConnectedToBoundary[rootX] |= self.isConnectedToBoundary[rootY]
            elif self.rank[rootX] < self.rank[rootY]:
                self.parent[rootX] = rootY
                self.isConnectedToBoundary[rootY] |= self.isConnectedToBoundary[rootX]
            else:
                self.parent[rootY] = rootX
                self.isConnectedToBoundary[rootX] |= self.isConnectedToBoundary[rootY]
                self.rank[rootX] += 1
    
    def setBoundaryConnection(self, x):
        root = self.find(x)
        self.isConnectedToBoundary[root] = True
    
    def isBoundaryConnected(self, x):
        return self.isConnectedToBoundary[self.find(x)]

class Solution:
    def numEnclaves(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        uf = UnionFind(m*n)

        def index(x, y):
            return x*n+y
        
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    if i == 0 or i == m - 1 or j == 0 or j == n - 1:
                        uf.setBoundaryConnection(index(i, j))
                    if i > 0 and grid[i - 1][j] == 1:
                        uf.union(index(i, j), index(i - 1, j))
                    if j > 0 and grid[i][j - 1] == 1:
                        uf.union(index(i, j), index(i, j - 1))

        count = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1 and not uf.isBoundaryConnected(index(i, j)):
                    count += 1

        return count
21403