<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:
        n=len(grid1)
        m=len(grid1[0])
        parent_set=set()
        grid_2=[]

        def dfs(r,c,child_set):
            if (r==-1 or r==n or c==-1 or c==m or (r,c) in parent_set or grid2[r][c]==0):
                return
            child_set.add((r,c))
            parent_set.add((r,c))
            dfs(r+1,c,child_set)
            dfs(r-1,c,child_set)
            dfs(r,c-1,child_set)
            dfs(r,c+1,child_set)

        for i in range(n):
            for j in range(m):
                if grid2[i][j]==1 and (i,j) not in parent_set:
                    child_set=set()
                    dfs(i,j,child_set)
                    if child_set:
                        grid_2.append(child_set)
        print(grid_2)
        count=0
        for island in grid_2:
            flag=1
            for r,c in island:
                if grid1[r][c]==0:
                    flag=0
                    break
            if flag==1:
                count+=1
        return count


        '''
        parent_set=set()
        def common(grid1,n,m,parent_set,grid_1):
            def dfs(r,c,child_set):
                if (r==-1 or r==n or c==-1 or c==m or (r,c) in parent_set or grid1[r][c]==0):
                    return 
                child_set.add((r,c))
                parent_set.add((r,c))
                dfs(r+1,c,child_set)
                dfs(r-1,c,child_set)
                dfs(r,c+1,child_set)
                dfs(r,c-1,child_set)
            for i in range(n):
                for j in range(m):
                    if grid1[i][j]==1 and (i,j) not in parent_set:
                        child_set=set()
                        dfs(i,j,child_set)
                        grid_1.append(list(child_set))
            #print(parent_set)
            return grid_1
        grid1_res=common(grid1,n,m,set(),[])
        grid2_res=common(grid2,n,m,set(),[])
        print(grid1_res)
        print("Space")
        print(grid2_res)
        count=0

        for island2 in grid2_res:
            is_sub_island = False
            for island1 in grid1_res:
                if set(island2).issubset(island1):
                    is_sub_island = True
                    break  
            if is_sub_island:
                count += 1
        return count
        '''




        
63070