<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 countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        m, n = len(grid1), len(grid1[0])
        def dfs(grid, i,j):
            if i < 0 or i >= m or j < 0 or j >=n or not grid[i][j] or (i,j) in visited: return
            visited.add((i,j))
            grid[i][j] = 0
            dfs(grid,i+1,j)
            dfs(grid, i-1,j)
            dfs(grid, i,j+1)
            dfs(grid, i,j-1)

        grid1_map = {}
        count = 0
        for i in range(m):
            for j in range(n):
                if grid1[i][j]:
                    visited = set()
                    dfs(grid1, i,j)
                    if len(visited) > 0: 
                        for node in visited:
                            grid1_map[node] = count
                        count+=1
        
        grid2_map = []
        for i in range(m):
            for j in range(n):
                if grid2[i][j]:
                    visited = set()
                    dfs(grid2, i,j)
                    grid2_map.append([node for node in visited])
        ans = 0
        val = 0
        # print(grid1_map)
        # print("      ----      ")
        # print(grid2_map)
        
        for island in grid2_map:
            flag = True
            for i in range(len(island)):
                if island[i] not in grid1_map: 
                    flag = False
                    break
                if i == 0: 
                    val = grid1_map[island[i]]
                else:
                    if val != grid1_map[island[i]]: 
                        flag = False
                        break
                    else:
                        flag = True
            if flag: ans +=1

        return ans
                



                


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