<p>You are given two <code>m x n</code> binary matrices <code>grid1</code> and <code>grid2</code> containing only <code>0</code>&#39;s (representing water) and <code>1</code>&#39;s (representing land). An <strong>island</strong> is a group of <code>1</code>&#39;s connected <strong>4-directionally</strong> (horizontal or vertical). Any cells outside of the grid are considered water cells.</p>

<p>An island in <code>grid2</code> is considered a <strong>sub-island </strong>if there is an island in <code>grid1</code> that contains <strong>all</strong> the cells that make up <strong>this</strong> island in <code>grid2</code>.</p>

<p>Return the <em><strong>number</strong> of islands in </em><code>grid2</code> <em>that are considered <strong>sub-islands</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/10/test1.png" style="width: 493px; height: 205px;" />
<pre>
<strong>Input:</strong> grid1 = [[1,1,1,0,0],[0,1,1,1,1],[0,0,0,0,0],[1,0,0,0,0],[1,1,0,1,1]], grid2 = [[1,1,1,0,0],[0,0,1,1,1],[0,1,0,0,0],[1,0,1,1,0],[0,1,0,1,0]]
<strong>Output:</strong> 3
<strong>Explanation: </strong>In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are three sub-islands.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/03/testcasex2.png" style="width: 491px; height: 201px;" />
<pre>
<strong>Input:</strong> grid1 = [[1,0,1,0,1],[1,1,1,1,1],[0,0,0,0,0],[1,1,1,1,1],[1,0,1,0,1]], grid2 = [[0,0,0,0,0],[1,1,1,1,1],[0,1,0,1,0],[0,1,0,1,0],[1,0,0,0,1]]
<strong>Output:</strong> 2 
<strong>Explanation: </strong>In the picture above, the grid on the left is grid1 and the grid on the right is grid2.
The 1s colored red in grid2 are those considered to be part of a sub-island. There are two sub-islands.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid1.length == grid2.length</code></li>
	<li><code>n == grid1[i].length == grid2[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>grid1[i][j]</code> and <code>grid2[i][j]</code> are either <code>0</code> or <code>1</code>.</li>
</ul>

class UnionFind:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.size = [1] * n

    def find(self, node):
        while node != self.parent[node]:
            self.parent[node] = self.parent[self.parent[node]]
            node = self.parent[node]
        return node

    def union(self, node1, node2):
        p1, p2 = self.find(node1), self.find(node2)
        if p1 == p2:
            return False

        if self.size[p1] > self.size[p2]:
            self.parent[p2] = p1
            self.size[p1] += self.size[p2]
        else:
            self.parent[p1] = p2
            self.size[p2] += self.size[p1]

        return True

class Solution:
    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        n, m = len(grid1), len(grid1[0])
        directions = [(-1, 0), (0, 1), (1, 0), (0, -1)]
        
        def index(x, y):
            return x * m + y
        
        uf = UnionFind(n * m)

        for i in range(n):
            for j in range(m):
                if grid2[i][j] == 1:
                    for dr, dc in directions:
                        r, c = i + dr, j + dc
                        if 0 <= r < n and 0 <= c < m and grid2[r][c] == 1:
                            uf.union(index(i, j), index(r, c))
        
        root_to_cells = defaultdict(list)
        for i in range(n):
            for j in range(m):
                if grid2[i][j] == 1:
                    root = uf.find(index(i, j))
                    root_to_cells[root].append((i, j))
        
        result = 0
        for cells in root_to_cells.values():
            if all(grid1[i][j] == 1 for i, j in cells):
                result += 1
        
        return result

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