<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>

from itertools import product
from collections import defaultdict
MVS = [(0, 1), (0, -1), (1, 0), (-1, 0)]
def isConn(grid):
    n, m = len(grid), len(grid[0])
    vis = set()
    def paint(x, y):
        for dx, dy in MVS:
            nx, ny = x + dx, y + dy
            if not (0 <= nx < n and 0 <= ny < m) or grid[nx][ny] == 0 or (nx, ny) in vis:
                continue
            vis.add((nx, ny))
            paint(nx, ny)
    pcs = 0
    for i, j in product(range(n), range(m)):
        if grid[i][j] == 0 or (i, j) in vis:
            continue
        pcs += 1
        paint(i, j)
    return pcs == 1

def hasAP(grid):
    n, m = len(grid), len(grid[0])
    time = 0
    disc = {}
    low = {}
    ap = False

    rx, ry = 0, 0
    for i, j in product(range(n), range(m)):
        if grid[i][j] == 1:
            rx, ry = i, j
            break
    assert(grid[rx][ry] == 1)
    def dfs(x, y):
        nonlocal ap, time
        time += 1
        disc[x, y] = time
        low[x, y] = time
        ch = 0
        for dx, dy in MVS:
            nx, ny = x + dx, y + dy
            if not (0 <= nx < n and 0 <= ny < m) or grid[nx][ny] == 0:
                continue
            if (nx, ny) not in disc:
                dfs(nx, ny)
                low[x, y] = min(low[x, y], low[nx, ny])
                ch += 1
                if (x, y) != (rx, ry) and low[nx, ny] >= disc[x, y]:
                    ap = True
            else:
                low[x, y] = min(low[x, y], disc[nx, ny])
        if (x, y) == (rx, ry) and ch > 1:
            ap = True
    dfs(rx, ry)
    return ap

class Solution:
    def minDays(self, grid: List[List[int]]) -> int:
        n, m = len(grid), len(grid[0])
        for i in range(n):
            print(grid[i])
        if not isConn(grid):
            return 0
        if sum(1 if grid[i][j] == 1 else 0 for i, j in product(range(n), range(m))) == 1 or hasAP(grid):
            return 1
        return 2
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