<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:
#         for i in range(len(grid2)):
#             for j in range(len(grid2[i])):
#                 if grid2[i][j] == 1 and grid1[i][j] == 0:
#                     grid2[i][j] = "X"
#                     contained = False
#                 elif grid2[i][j] == 1 and grid1[i][j] == 1:
#                     grid2[i][j] = "[]"
#         # grid2, regions = self.infect_regions_and_count(grid2)
#         return self.infect_regions_and_count(grid2)
    
#     def infect_regions_and_count(self, grid):
#         region_number = 1
#         water = 0
#         # make a top border
#         new_grid = [[water for _ in range(len(grid[0])+2)]]
#         # fill in the grid with left and right border
#         for i in range(len(grid)):
#             new_grid.append([water]+grid[i]+[water])
#         # make a bottom border
#         new_grid.append([water for _ in range(len(grid[0])+2)])
#         for i in range(1,len(new_grid)-1):
#             for j in range(1, len(new_grid[i])-1):
#                 if new_grid[i][j] == "X":
#                     self.flood_fill([i,j], new_grid, 0)
#         region_number = 1
#         for i in range(1,len(new_grid)-1):
#             for j in range(1, len(new_grid[i])-1):
#                 if new_grid[i][j] == "[]":
#                     self.flood_fill([i,j], new_grid, region_number)
#                     region_number+=1
#         # return new_grid, region_number
#         return region_number-1
    
#     def flood_fill(self, node, new_grid, region_number):
#         direction = [[-1,0],[0,-1],[1,0],[0,1]]
#         Q = []
#         Q += [node]
#         while Q:
#             current_node = Q[0]
#             if current_node[0] < 0 or current_node[1] < 0 or current_node[0] > len(new_grid)-1 or current_node[1] > len(new_grid[0])-1:
#                 print("error")
#             if new_grid[current_node[0]][current_node[1]] == "[]" or new_grid[current_node[0]][current_node[1]]=="X":
#                 new_grid[current_node[0]][current_node[1]] = region_number
#                 Q += [[current_node[0] + direction[0][0], current_node[1] + direction[0][1]]]
#                 Q += [[current_node[0] + direction[1][0], current_node[1] + direction[1][1]]]
#                 Q += [[current_node[0] + direction[2][0], current_node[1] + direction[2][1]]]
#                 Q += [[current_node[0] + direction[3][0], current_node[1] + direction[3][1]]]
#             Q.pop(0)

# class Solution:
#     # Directions in which we can traverse inside the grids.
#     directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]

#     # Helper method to check if the cell at the position (x, y) in the 'grid'
#     # is a land cell.
#     def is_cell_land(self, x, y, grid):
#         return grid[x][y] == 1

#     # Union-Find class.
#     class UnionFind:
#         def __init__(self, n):
#             self.parent = list(range(n))
#             self.rank = [0] * n

#         # Find the root of element 'u', using the path-compression technique.
#         def find(self, u):
#             if self.parent[u] != u:
#                 self.parent[u] = self.find(self.parent[u])
#             return self.parent[u]

#         # Union two components of elements 'u' and 'v' respectively based on their ranks.
#         def union_sets(self, u, v):
#             root_u = self.find(u)
#             root_v = self.find(v)
#             if root_u != root_v:
#                 if self.rank[root_u] > self.rank[root_v]:
#                     self.parent[root_v] = root_u
#                 elif self.rank[root_u] < self.rank[root_v]:
#                     self.parent[root_u] = root_v
#                 else:
#                     self.parent[root_v] = root_u
#                     self.rank[root_u] += 1

#     # Helper method to convert (x, y) position to a 1-dimensional index.
#     def convert_to_index(self, x, y, total_cols):
#         return x * total_cols + y

#     def countSubIslands(
#         self, grid1: List[List[int]], grid2: List[List[int]]
#     ) -> int:
#         total_rows = len(grid2)
#         total_cols = len(grid2[0])
#         uf = self.UnionFind(total_rows * total_cols)

#         # Traverse each land cell of 'grid2'.
#         for x in range(total_rows):
#             for y in range(total_cols):
#                 if self.is_cell_land(x, y, grid2):
#                     # Union adjacent land cells with the current land cell.
#                     for direction in self.directions:
#                         next_x = x + direction[0]
#                         next_y = y + direction[1]
#                         if (
#                             0 <= next_x < total_rows
#                             and 0 <= next_y < total_cols
#                             and self.is_cell_land(next_x, next_y, grid2)
#                         ):
#                             uf.union_sets(
#                                 self.convert_to_index(x, y, total_cols),
#                                 self.convert_to_index(
#                                     next_x, next_y, total_cols
#                                 ),
#                             )

#         # Traverse 'grid2' land cells and mark that cell's root as not a sub-island
#         # if the land cell is not present at the respective position in 'grid1'.
#         is_sub_island = [True] * (total_rows * total_cols)
#         for x in range(total_rows):
#             for y in range(total_cols):
#                 if self.is_cell_land(x, y, grid2) and not self.is_cell_land(
#                     x, y, grid1
#                 ):
#                     root = uf.find(self.convert_to_index(x, y, total_cols))
#                     is_sub_island[root] = False

#         # Count all the sub-islands.
#         sub_island_counts = 0
#         for x in range(total_rows):
#             for y in range(total_cols):
#                 if self.is_cell_land(x, y, grid2):
#                     root = uf.find(self.convert_to_index(x, y, total_cols))
#                     if is_sub_island[root]:
#                         sub_island_counts += 1
#                         # One cell can be the root of multiple land cells, so to
#                         # avoid counting the same island multiple times, mark it as false.
#                         is_sub_island[root] = False

        # return sub_island_counts

class Solution:
    def countSubIslands(self, grid1: List[List[int]], grid2: List[List[int]]) -> int:
        self.grid1 = grid1
        self.grid2 = grid2
        self.rownum = len(grid2)
        self.colnum = len(grid2[0])
        return self.count_subislands()
    
    def count_subislands(self) -> int:
        count = 0
        for i in range(self.rownum):
            for j in range(self.colnum):
                if self.grid2[i][j] == 1:
                    self.valid_island = True
                    self.process_island(i, j)
                    if self.valid_island:
                        count += 1
                    self.valid_island = True
        return count
    
    def process_island(self, i: int, j: int) -> None:
        self.grid2[i][j] = 0
        if self.grid1[i][j] == 0:
            self.valid_island = False

        if i > 0 and self.grid2[i - 1][j] == 1:
            self.process_island(i - 1, j)
        if i < self.rownum - 1 and self.grid2[i + 1][j] == 1:
            self.process_island(i + 1, j)
        if j > 0 and self.grid2[i][j - 1] == 1:
            self.process_island(i, j - 1)
        if j < self.colnum - 1 and self.grid2[i][j + 1] == 1:
            self.process_island(i, j + 1)
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