<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 typing import List
from collections import deque

class Solution:
    def minDays(self, grid: List[List[int]]) -> int:
        n, m = len(grid), len(grid[0])

        land = sum([sum(grid[i]) for i in range(len(grid))])

        def island_size(x, y):
            grid_copy = [grid[i][:] for i in range(n)]
            grid_copy[x][y] = 0
            d = deque([[x, y]])
            s = 1

            while len(d) > 0:
                x_, y_ = d.popleft()

                if 0 <= y_ < m:
                    if x_ > 0 and grid_copy[x_ - 1][y_] == 1:
                        grid_copy[x_ - 1][y_] = 0
                        d.append([x_ - 1, y_])
                        s += 1

                    if x_ < n - 1 and grid_copy[x_ + 1][y_] == 1:
                        grid_copy[x_ + 1][y_] = 0
                        d.append([x_ + 1, y_])
                        s += 1

                if 0 <= x_ < n:
                    if y_ > 0 and grid_copy[x_][y_ - 1] == 1:
                        grid_copy[x_][y_ - 1] = 0
                        d.append([x_, y_ - 1])
                        s += 1

                    if y_ < m - 1 and grid_copy[x_][y_ + 1] == 1:
                        grid_copy[x_][y_ + 1] = 0
                        d.append([x_, y_ + 1])
                        s += 1

            return s

        # def is_bridge(x, y):
        #     if 0 < x < n - 1:
        #         if grid[x - 1][y] == 1 and grid[x + 1][y] == 1:
        #             if y == 0:


        def find_land():
            find = False
            for i in range(n):
                for j in range(m):
                    if grid[i][j] != 1:
                        continue

                    find = True
                    break

                if find:
                    break

            return i, j
        
        if land == 0:
            return 0
        
        i, j = find_land()

        s = island_size(i, j)
        if s < land:
            return 0
        
        if s == 1:
            return 1
        
        for i in range(n):
            for j in range(m):
                if grid[i][j] == 0:
                    continue

                if True:
                    grid[i][j] = 0

                    i_, j_ = find_land()
                    s_ = island_size(i_, j_)
                    if s_ + 1 < land:
                        return 1
                    
                    grid[i][j] = 1

        return 2

# print(Solution().minDays(grid = [[0,1,1,0],[0,1,1,0],[0,0,0,0]]), 2)
# print(Solution().minDays(grid = [[1,1]]), 2)
# print(Solution().minDays([[1,1,0,1,1],[1,1,1,1,1],[1,1,0,1,1],[1,1,0,1,1]]), 1)
# print(Solution().minDays([[1,1,1,1],[1,1,1,0],[1,1,0,1],[1,1,1,1],[1,1,0,0],[1,1,1,1]]), 1)
print(Solution().minDays([[0,0]]), 0)

414