<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:
        COLS = len(grid[0])
        ROWS = len(grid)
        def count_island(grid):
            visit = set()
            def dfs(r,c):
                visit.add((r,c))
                for i, j in [(-1,0), (1,0), (0,1), (0, -1)]:
                    nxt_r = r + i
                    nxt_c = c + j
                    if 0 <= nxt_r < ROWS and 0 <= nxt_c < COLS and grid[nxt_r][nxt_c] == 1 and (nxt_r, nxt_c) not in visit:
                        dfs(nxt_r, nxt_c)
            count = 0
            for r in range(ROWS):
                for c in range(COLS):
                    if grid[r][c] == 1 and (r,c) not in visit:
                        count += 1
                        dfs(r, c)
            return count
        initial_count = count_island(grid)
        print(initial_count)
        if initial_count != 1:
            return 0
        # check if we can find a bridge or not
        def get_index(r, c):
            return r * COLS + c
        
        for r in range(ROWS):
            for c in range(COLS):
                if grid[r][c] == 1:
                    grid[r][c] = 0
                    count = count_island(grid)
                    # print(r,c ,"count:", count)
                    if count != 1: return 1
                    grid[r][c] = 1
        return 2

        
        # def get_bridges(grid):
        #     graph = defaultdict(list)
        #     for r in range(ROWS):
        #         for c in range(COLS):
        #             if grid[r][c] == 0: continue
        #             for i,j in [(-1,0),(1,0),(0,-1),(0,1)]:
        #                 nxt_r = r + i
        #                 nxt_c = c + i
        #                 if 0 <= nxt_r < ROWS and 0 <= nxt_c < COLS and grid[nxt_r][nxt_c] == 1:
        #                     graph[get_index(r,c)] = get_index(nxt_r,nxt_c)
        #                     graph[get_index(nxt_r,nxt_c)] = get_index(r,c)
        #     # Tarjan algorithm to find bridge
        #     dist = [-1] * len(graph)
        #     low = [-1] * len(graph)
        #     visit = set()
        #     self.time = 0 
        #     bridges = []
        #     def dfs(prev, cur):
        #         visit.add(cur)
        #         self.time += 1
        #         dist[cur], low[cur] = self.time, self.time
        #         for nxt in graph[cur]:
        #             if nxt not in visit:
        #                 # creating a spanning tree
        #                 dfs(cur, nxt)
        #                 low[cur] = min(low[cur], low[nxt])
        #                 if dist[cur] < low[nxt]:
        #                     bridges.append([cur, nxt])
        #             else:
        #                 low[cur] = min(low[cur], dist[nxt])
        #     return bridges
        
        # bridges = get_bridges(grid)
        # if bridges: return 1
        # else:
        #     return 2
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