<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 DFS
    def dfs(self, grid, visited, x, y):
        rows, cols = len(grid), len(grid[0])
        visited[x][y] = True
        
        # Define the directions for moving in the grid (up, down, left, right)
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
        
        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < rows and 0 <= ny < cols and not visited[nx][ny] and grid[nx][ny] == 1:
                self.dfs(grid, visited, nx, ny)
    
    # Function to count the number of connected components (islands)
    def countComponents(self, grid):
        rows, cols = len(grid), len(grid[0])
        visited = [[False] * cols for _ in range(rows)]
        count = 0
        
        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 1 and not visited[i][j]:
                    self.dfs(grid, visited, i, j)
                    count += 1
        return count
    
    # Function to determine the minimum number of days to disconnect the island
    def minDays(self, grid) -> int:
        rows, cols = len(grid), len(grid[0])
        
        # Step 1: Check if the island is already disconnected
        if self.countComponents(grid) != 1:
            return 0
        
        # Step 2: Try removing each land cell and check if it disconnects the island
        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 1:
                    # Temporarily set the current cell to water
                    grid[i][j] = 0
                    if self.countComponents(grid) != 1:  # Check if it disconnects the island
                        return 1
                    # Revert the change
                    grid[i][j] = 1
        
        # Step 3: If one removal doesn't work, two removals are always sufficient
        return 2

# Example usage
sol = Solution()
grid = [
    [1, 1, 1, 0, 0],
    [1, 1, 0, 0, 1],
    [0, 1, 0, 1, 1],
    [0, 0, 0, 1, 1]
]
print(sol.minDays(grid))  # Output: 2

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