<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 __init__(self):
        self.isSubIsland = True

    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        rowL = len(grid1)
        colL = len(grid1[0])
        visited = [[False] * colL for _ in range(rowL)]
        subIslands = 0
        for row in range(rowL):
            for col in range(colL):
                if visited[row][col] or grid2[row][col] == 0:
                    continue
                
                self.isSubIsland = True
                self.dfs(grid1, grid2, row, col, rowL, colL, visited)
                if self.isSubIsland:
                    subIslands += 1
        
        return subIslands
    
    def dfs(self, grid1, grid2, row, col, rowL, colL, visited):
        if visited[row][col] or grid2[row][col] == 0:
            return
        
        visited[row][col] = True
        if grid2[row][col] != grid1[row][col]:
            self.isSubIsland = False
        
        for nRow, nCol in self.getNeighbors(grid1, row, col, rowL, colL):
            self.dfs(grid1, grid2, nRow, nCol, rowL, colL, visited)
    
    def getNeighbors(self, grid, row, col, rowL, colL):
        neighbors = []
        if row > 0:
            neighbors.append((row - 1, col))
        if col > 0:
            neighbors.append((row, col -1))
        if row < rowL - 1:
            neighbors.append((row + 1, col))
        if col < colL - 1:
            neighbors.append((row, col + 1))
        
        return neighbors

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