<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

    # Traverse all cells of island starting at position (x, y) in 'grid2',
    # and check this island is a sub-island in 'grid1'.
    def is_sub_island(self, x, y, grid1, grid2, visited):
        total_rows = len(grid2)
        total_cols = len(grid2[0])

        is_sub_island = True

        pending_cells = deque()
        # Push the starting cell in the queue and mark it as visited.
        pending_cells.append((x, y))
        visited[x][y] = True

        # Traverse on all cells using the breadth-first search method.
        while pending_cells:
            curr_x, curr_y = pending_cells.popleft()

            # If the current position cell is not a land cell in 'grid1',
            # then the current island can't be a sub-island.
            if not self.is_cell_land(curr_x, curr_y, grid1):
                is_sub_island = False

            for direction in self.directions:
                next_x = curr_x + direction[0]
                next_y = curr_y + direction[1]
                # If the next cell is inside 'grid2', is never visited and
                # is a land cell, then we traverse to the next cell.
                if (
                    0 <= next_x < total_rows
                    and 0 <= next_y < total_cols
                    and not visited[next_x][next_y]
                    and self.is_cell_land(next_x, next_y, grid2)
                ):
                    # Push the next cell in the queue and mark it as visited.
                    pending_cells.append((next_x, next_y))
                    visited[next_x][next_y] = True
        return is_sub_island

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

        visited = [[False] * total_cols for _ in range(total_rows)]
        sub_island_counts = 0

        # Iterate on each cell in 'grid2'
        for x in range(total_rows):
            for y in range(total_cols):
                # If cell at the position (x, y) in the 'grid2' is not visited,
                # is a land cell in 'grid2', and the island
                # starting from this cell is a sub-island in 'grid1', then we
                # increment the count of sub-islands.
                if (
                    not visited[x][y]
                    and self.is_cell_land(x, y, grid2)
                    and self.is_sub_island(x, y, grid1, grid2, visited)
                ):
                    sub_island_counts += 1

        # Return total count of sub-islands.
        return sub_island_counts
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