<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:
        # must first count how many islands there are using dfs and set up edge count
        def dfs(i: int, j: int) -> None:
            directions = [(-1, 0), (1, 0), (0, 1), (0, -1)]
            for dx, dy in directions:
                i2, j2 = i + dx, j + dy
                if is_in_bounds(i2, j2) and grid[i2][j2] == 1 and (i2, j2) not in visited:
                    visited.add((i2, j2))
                    dfs(i2, j2)
        
        def is_in_bounds(i: int, j: int) -> bool:
            return 0 <= i < m and 0 <= j < n
        
        # Need to know degree for each node -> create a map of visited_to_degree and it's min values
        # need to know number of islands -> num_comp
        # need to know number of squares in the island -> len(visited) if num_comp == 1
        m, n = len(grid), len(grid[0])
        visited = set()
        num_components = 0
        for r in range(m):
            for c in range(n):
                if grid[r][c] == 1 and (r, c) not in visited:
                    num_components += 1
                    visited.add((r, c))
                    dfs(r, c)
        
        if num_components != 1:
            return 0 # Already disconnected

        # min_degree = min(visited_to_deg.values())
        # if len(visited_to_deg) == min_degree + 1:
        #     return min_degree + 1 # Case where all have to be deleted
        # return min_degree
        # NOTE: I was totally wrong. using hints i'll now finish the problem by checking if deleting an edge
        # makes num_components == 2
        visited_lst = list(visited)
        if len(visited_lst) == 1:
            return 1
        for i in range(len(visited_lst)):
            j = (i + 1) % len(visited_lst)
            r, c = visited_lst[i]
            r2, c2 = visited_lst[j]
            visited = set([(r2, c2)])
            grid[r][c] = 0 # Test as zero
            dfs(r2, c2) # Should visit len(visited) - 1 starting from another node
            if len(visited) < len(visited_lst) - 1:
                return 1
            grid[r][c] = 1
        return 2

        
        # 
        # 1 1 1 1 1
        # 1 1 1 1 1
        # 1 1 1 1 1
        # 1 1 1 1 1


1419