<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:
    def getIslands(self, grid):
        M, N = len(grid), len(grid[0])
        islands = []
        seen = set()
        neighbors = [
            (0, 1), (0, -1), (1, 0), (-1, 0)
        ]
        
        def is_valid_neighbor(x, y):
            return 0 <= x < M and 0 <= y < N and grid[x][y] == 1 and (x, y) not in seen
        
        for m in range(M):
            for n in range(N):
                if grid[m][n] == 1 and (m, n) not in seen:
                    # do the dfs
                    island = []
                    stack = [(m, n)]
                    seen.add((m, n))
                    while stack:
                        cm, cn = stack.pop()
                        island.append((cm, cn))
                        for dm, dn in neighbors:
                            mx, nx = cm + dm, cn + dn
                            if is_valid_neighbor(mx, nx):
                                stack.append((mx, nx))
                                seen.add((mx, nx))
                        
                    islands.append(island)
        
        return islands
                        
    
    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        islands = self.getIslands(grid2)
        count = 0
        
        for island in islands:
            is_sub_island = True
            for m, n in island:
                if grid1[m][n] != 1:
                    is_sub_island = False
                    
            if is_sub_island:
                count += 1
                
        return count
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