<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: List[List[int]]) -> int:
        # https://leetcode.com/u/magnetize/

        def depthMark (grid, row, col):
            if grid[row][col] == 0 or grid[row][col] == -1:
                return

            grid[row][col] = -1
            if row > 0:
                depthMark(grid, row - 1, col)
            if col > 0:
                depthMark(grid, row, col - 1)
            if row < len(grid) - 1:
                depthMark(grid, row + 1, col)
            if col < len(grid[0]) - 1:
                depthMark(grid, row, col + 1)
            return            

        def numIsland (grid) -> int:

            islands = 0
            for row in range(len(grid)):
                for col in range(len(grid[0])):
                    if grid[row][col] == 1:
                        islands += 1
                        if islands >= 2:
                            break
                        depthMark(grid, row, col)

            # Restore the grid to original.
            for row in range(len(grid)):
                for col in range(len(grid[0])):
                    if grid[row][col] == -1:
                        grid[row][col] = 1
            return islands

        # If the original grid has 0 islands or 2 or more islands then no changes are needed.
        islands = numIsland (grid)
        if islands == 0 or islands >= 2:
            return 0

        # Try changing one cell and check if we can get two islands.
        for row in range(len(grid)):
            for col in range(len(grid[0])):
                if grid[row][col] == 1:
                    grid[row][col] = 0
                    islands = numIsland (grid)
                    grid[row][col] = 1
                    if islands == 0 or islands >= 2:
                        return 1

        # We can't generate two islands using a single update, so we must need two.
        return 2
585