<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:
        R, C = len(grid), len(grid[0])

        def dfs(r, c, visited):
            if (r, c) in visited:
                return
            visited.add((r, c))
            for dx, dy in [[0, 1], [1, 0], [-1, 0], [0, -1]]:
                nextr, nextc = r+dx, c+dy
                if nextr < 0 or nextr >=R or nextc < 0 or nextc >= C:
                    continue
                if grid[nextr][nextc] == 1:
                    dfs(nextr, nextc, visited)
                
        # 1. find nr of islands, if we got 0 or >1 land, means we already have
        # disconnected islands
        islands_count = 0
        visited = set()
        for r in range(R):
            for c in range(C):
                if grid[r][c] == 1 and (r, c) not in visited:
                    dfs(r, c, visited)
                    islands_count+=1

        if islands_count == 0 or islands_count > 1:
            # either no islands or more than one islands found
            return 0 # no cell need to turn into water
        
        # 2. count the islands again
        land = list(visited)
        for r, c in land:
             # try turning each land cell to water and count the islands again
             # if we get more than one islands, then we need one day
            grid[r][c] = 0
            islands_count = 0
            visited = set()
            for x in range(R):
                for y in range(C):
                    if grid[x][y] == 1 and (x, y) not in visited:
                        dfs(x, y, visited)
                        islands_count+=1

            if islands_count == 0 or islands_count > 1:
                return 1
            
            grid[r][c] = 1

        # else we need 2 days to convert any 2 cell to water
        return 2
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