<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)]

    def is_cell_land(self, x, y, grid):
        return grid[x][y] == 1

    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()
        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 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 (
                    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)
                ):
                    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 (
                    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 sub_island_counts
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