<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:
        # an island in grid2 is considered a sub-island if each of its cells exist within an island in grid1
        # graph
        # if stack is empty, then count +=1
        # change visited cells to 2
        # create helper function to add neighbors to stack 

        # this will help us find out how many islands exist in graph2, but we need to find overlapping islands
            # [0,5] =1?, how many islands would we have? 0
            # verify sub-island coordinates
            # recursive helper function collects island coordinates, after stack is empty, verify coordinates in graph1 using a for loop
                # only increase count if there is an overlap for each coordinate
        
        count = 0
        dir_arr = [[-1,0], [1,0], [0,1], [0,-1]]
        length = len(grid1)
        width = len(grid1[0])

        def dfs(coords, stack):
            # recursive loop
            while stack:
                # print("coords: ", coords)
                i,j = stack.pop()
                for x,y in dir_arr:
                    new_x, new_y = x+i, y+j
                    if 0<= new_x < length and 0<= new_y < width and grid2[new_x][new_y] == 1:
                        stack.append([new_x, new_y])
                        coords.append([new_x, new_y])
                        grid2[new_x][new_y] = 2


            # update island?
            for i,j in coords:
                if grid1[i][j] != 1:
                    return False
            return True
            # print("count: ", count)
            
        for i in range(length):
            for j in range(width):
                if grid2[i][j] == 1:
                    coords = [[i,j]]
                    stack = [[i,j]]
                    if dfs(coords, stack):
                        count +=1
        return count


        
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