<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>

from typing import List

class Solution:
    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        m, n = len(grid1), len(grid1[0])
        count = 0

        def dfs(r, c):
            if not (0 <= r < m and 0 <= c < n) or grid2[r][c] != 1:
                return True
            if grid1[r][c] != 1:
                return False

            grid2[r][c] = -1
            result = True
            for dr, dc in [(0, 1), (1, 0), (-1, 0), (0, -1)]:
                result = dfs(r + dr, c + dc) and result
            return result

        for i in range(m):
            for j in range(n):
                if grid2[i][j] == 1:
                    if dfs(i, j):
                        count += 1

        return count

# class Solution:
#     def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
#         m = len(grid1)
#         n = len(grid1[0])
#         count = 0
#         # 1 if all neighbors are also satisyfing tule in grid1
#             # else 0
#         def dfs(r,c):
#             grid2[r][c] = "-1"
#             dirs = [(0, 1), (1, 0), (-1,0), (0, -1)]
#             for dr, dc in dirs:
#                 nr = r+dr
#                 nc = c+dc
#                 if 0<=nr<m and 0<=nc<n and grid2[nr][nc]==1 and grid1[nr][nc]==1:
#                     print("go deep at", nr, nc)
#                     dfs(nr, nc)
#                 elif 0<=nr<m and 0<=nc<n and (grid2[nr][nc]!=1 or grid1[nr][nc]!=1):
#                     print("quit at", nr, nc)
#                     return 0
#             return 1
#         for i in range(m):
#             for j in range(n):
#                 if grid2[i][j]==1 and grid1[i][j]==1:
#                     print("enter dfs at", i, j)
#                     count += dfs(i, j)
#         return count
        
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