<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:
    DIRS = [[0, 1], [1, 0], [0, -1], [-1, 0]]

    def minDays(self, grid: List[List[int]]) -> int:
        rows, cols = len(grid), len(grid[0])
        apPointFound = False
        apPointTime = 0

        landCells = 0
        islandCount = 0

        discoveryTime = [[-1] * cols for _ in range(rows)]
        lowestReachable = [[-1] * cols for _ in range(rows)]
        parentCell = [[-1] * cols for _ in range(rows)]

        def findArticulationPoint(r, c):
            nonlocal apPointFound, apPointTime
            discoveryTime[r][c] = apPointTime
            apPointTime += 1
            lowestReachable[r][c] = discoveryTime[r][c]
            children = 0

            for dir in self.DIRS:
                nr = r + dir[0]
                nc = c + dir[1]

                if 0<=nr<rows and 0<=nc<cols and grid[nr][nc]==1:
                    if discoveryTime[nr][nc] == -1:
                        children += 1
                        parentCell[nr][nc] = r * cols + c
                        findArticulationPoint(nr, nc)

                        lowestReachable[r][c] = min(
                            lowestReachable[r][c],
                            lowestReachable[nr][nc],
                        )

                        if lowestReachable[nr][nc] >= discoveryTime[r][c] and parentCell[r][c] != -1:
                            apPointFound = True
                    elif nr*cols + nc != parentCell[r][c]:
                        lowestReachable[r][c] = min(
                            lowestReachable[r][c],
                            discoveryTime[nr][nc],
                        )

            if parentCell[r][c] == -1 and children > 1:
                apPointFound = True
        
        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 1:
                    landCells += 1
                    if discoveryTime[i][j] == -1:
                        findArticulationPoint(i, j)
                        islandCount += 1

        # print(f'-- {islandCount} -- {landCells} -- {apPointFound}')
        if islandCount == 0 or islandCount >= 2:
            return 0
        if landCells == 1:
            return 1
        if apPointFound:
            return 1
        return 2





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