<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:
    # Helper function to perform Depth-First Search (DFS) to explore the island
    def dfs(self, grid: List[List[int]], visited: List[List[bool]], i: int, j: int):
        m, n = len(grid), len(grid[0])
        # Check boundaries and whether the cell is already visited or is water
        if i < 0 or j < 0 or i >= m or j >= n or grid[i][j] == 0 or visited[i][j]:
            return
        visited[i][j] = True # Mark the cell as visited
        # Recursively visit all 4 directions
        self.dfs(grid, visited, i+1, j)
        self.dfs(grid, visited, i-1, j)
        self.dfs(grid, visited, i, j+1)
        self.dfs(grid, visited, i, j-1)

    # Function to count the number of islands (connected groups of 1's)
    def countIslands(self, grid: List[List[int]]) -> int:
        m, n = len(grid), len(grid[0])
        visited = [[False] * n for _ in range(m)]
        count = 0

        # Iterate through the grid to find and count each island
        for i in range(m):
            for j in range(n):
                # If we find a land cell that is not visited, start a new DFS
                if grid[i][j] == 1 and not visited[i][j]:
                    count += 1
                    self.dfs(grid, visited, i, j) # Explore the island
        return count
    
    # Function to determine the minimum number of days to disconnect the island
    def minDays(self, grid: List[List[int]]) -> int:
        # Step 1: Check if the grid is already disconnected
        if self.countIslands(grid) != 1:
            return 0
        
        m, n = len(grid), len(grid[0])

        # Step 2: Try removing one land cell
        for i in range(m):
            for j in range(n):
                if grid[i][j] == 1:
                    grid[i][j] = 0 # Remove the land cell
                    # Check if removing this cell disconnects the grid
                    if self.countIslands(grid) != 1:
                        return 1
                    grid[i][j] = 1 # Restore the land cell

        # Step 3: If removing one cell isn't enough, return 2 (as it is guaranteed to be sufficient)
        return 2
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