<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 Solution:
    # Directions in which we can traverse inside the grids.
    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]

    # Helper method to check if the cell at the position (x, y) in the 'grid'
    # is a land cell.
    def is_cell_land(self, x, y, grid):
        return grid[x][y] == 1

    # Union-Find class.
    class UnionFind:
        def __init__(self, n):
            self.parent = list(range(n))
            self.rank = [0] * n

        # Find the root of element 'u', using the path-compression technique.
        def find(self, u):
            if self.parent[u] != u:
                self.parent[u] = self.find(self.parent[u])
            return self.parent[u]

        # Union two components of elements 'u' and 'v' respectively based on their ranks.
        def union_sets(self, u, v):
            root_u = self.find(u)
            root_v = self.find(v)
            if root_u != root_v:
                if self.rank[root_u] > self.rank[root_v]:
                    self.parent[root_v] = root_u
                elif self.rank[root_u] < self.rank[root_v]:
                    self.parent[root_u] = root_v
                else:
                    self.parent[root_v] = root_u
                    self.rank[root_u] += 1

    # Helper method to convert (x, y) position to a 1-dimensional index.
    def convert_to_index(self, x, y, total_cols):
        return x * total_cols + y

    def countSubIslands(
        self, grid1: List[List[int]], grid2: List[List[int]]
    ) -> int:
        total_rows = len(grid2)
        total_cols = len(grid2[0])
        uf = self.UnionFind(total_rows * total_cols)

        # Traverse each land cell of 'grid2'.
        for x in range(total_rows):
            for y in range(total_cols):
                if self.is_cell_land(x, y, grid2):
                    # Union adjacent land cells with the current land cell.
                    for direction in self.directions:
                        next_x = x + direction[0]
                        next_y = y + direction[1]
                        if (
                            0 <= next_x < total_rows
                            and 0 <= next_y < total_cols
                            and self.is_cell_land(next_x, next_y, grid2)
                        ):
                            uf.union_sets(
                                self.convert_to_index(x, y, total_cols),
                                self.convert_to_index(
                                    next_x, next_y, total_cols
                                ),
                            )

        # Traverse 'grid2' land cells and mark that cell's root as not a sub-island
        # if the land cell is not present at the respective position in 'grid1'.
        is_sub_island = [True] * (total_rows * total_cols)
        for x in range(total_rows):
            for y in range(total_cols):
                if self.is_cell_land(x, y, grid2) and not self.is_cell_land(
                    x, y, grid1
                ):
                    root = uf.find(self.convert_to_index(x, y, total_cols))
                    is_sub_island[root] = False

        # Count all the sub-islands.
        sub_island_counts = 0
        for x in range(total_rows):
            for y in range(total_cols):
                if self.is_cell_land(x, y, grid2):
                    root = uf.find(self.convert_to_index(x, y, total_cols))
                    if is_sub_island[root]:
                        sub_island_counts += 1
                        # One cell can be the root of multiple land cells, so to
                        # avoid counting the same island multiple times, mark it as false.
                        is_sub_island[root] = False

        return sub_island_counts
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