<p>You are given an <code>m x n</code> binary grid <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water. An <strong>island</strong> is a maximal <strong>4-directionally</strong> (horizontal or vertical) connected group of <code>1</code>&#39;s.</p>

<p>The grid is said to be <strong>connected</strong> if we have <strong>exactly one island</strong>, otherwise is said <strong>disconnected</strong>.</p>

<p>In one day, we are allowed to change <strong>any </strong>single land cell <code>(1)</code> into a water cell <code>(0)</code>.</p>

<p>Return <em>the minimum number of days to disconnect the grid</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land1.jpg" style="width: 500px; height: 169px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]

<strong>Output:</strong> 2
<strong>Explanation:</strong> We need at least 2 days to get a disconnected grid.
Change land grid[1][1] and grid[0][2] to water and get 2 disconnected island.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/12/24/land2.jpg" style="width: 404px; height: 85px;" />
<pre>
<strong>Input:</strong> grid = [[1,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Grid of full water is also disconnected ([[1,1]] -&gt; [[0,0]]), 0 islands.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 30</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution:
    def minDays(self, grid):
        m,n = len(grid),len(grid[0])
        # Arrays to store information for each cell
        # Time when a cell is first discovered
        discover = [[-1]*n for _ in range(m)]
        # Lowest discovery time reachable from the subtree rooted at this cell
        low = [[-1]*n for _ in range(m)]
        # Parent of each cell in DFS tree
        parent = [[-1]*n for _ in range(m)]
        time = 0
        isArticulatePresent = False
        directions = [[-1,0],[1,0],[0,-1],[0,1]]
        land_cells = island_count = 0

        def dfs(row,col):
            nonlocal time,m,n,isArticulatePresent
            low[row][col] = discover[row][col] = time
            time+=1
            children = 0
            
            # Explore all adjacent cells
            for dx,dy in directions:
                nx,ny = row+dx,col+dy
                if 0<=nx<m and 0<=ny<n and grid[nx][ny] == 1:
                    if nx*n+ny == parent[row][col]:continue
                    elif discover[nx][ny]!=-1: low[row][col] = min(low[row][col],discover[nx][ny])
                    else:
                        children+=1
                        # Set parent
                        parent[nx][ny] = row*n+col
                        dfs(nx,ny)
                        # Update lowest reachable time
                        low[row][col] = min(low[row][col],low[nx][ny])
                         # Check for articulation point condition
                        if low[nx][ny]>=discover[row][col] and parent[row][col]!=-1: isArticulatePresent = True
            # Root of DFS tree is an articulation point if it has more than one child
            if parent[row][col] == -1 and children > 1: isArticulatePresent = True

        # Traverse the grid to find islands and articulation points
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    land_cells+=1
                    if discover[i][j] == -1:# If not yet visited
                        # Start DFS for a new island
                        dfs(i,j)
                        island_count += 1

        # Determine the minimum number of days to disconnect the grid
        if island_count == 0 or island_count >= 2: return 0  # Already disconnected or no land
        if land_cells == 1: return 1  # Only one land cell
        if isArticulatePresent: return 1  # An articulation point exists
        return 2  # Need to remove any two land cells

# class Solution:
#     def minDays(self, grid: List[List[int]]) -> int:
#         m,n = len(grid),len(grid[0])
#         directions = [[-1,0],[1,0],[0,-1],[0,1]]
        
#         def islandCount():
#             count,visited = 0,set()
#             for i in range(m):
#                 for j in range(n):
#                     if grid[i][j] == 1 and (i,j) not in visited:
#                         count+=1
#                         visited.add((i,j))
#                         queue = deque([[i,j]])
#                         while queue:
#                             x,y = queue.popleft()
#                             for dx,dy in directions:
#                                 nx,ny = x+dx,y+dy
#                                 if 0<=nx<m and 0<=ny<n and (nx,ny) not in visited and grid[nx][ny] == 1:
#                                     visited.add((nx,ny))
#                                     queue.append([nx,ny])
#             return count                        
        
#         res = islandCount()
#         if res>1 or res == 0: return 0

#         for i in range(m):
#             for j in range(n):
#                 if grid[i][j] == 1:
#                     grid[i][j] = 0
#                     res = islandCount()
#                     if res>1 or res == 0: return 1
#                     grid[i][j] = 1

#         return 2                                 
        
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