Minimum Number of Days to Disconnect Island
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1568.Minimum%20Number%20of%20Days%20to%20Disconnect%20Island/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1500-1599/1568.Minimum%20Number%20of%20Days%20to%20Disconnect%20Island/images/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>
Depth-First Search; Breadth-First Search; Array; Matrix; Strongly Connected Component
func minDays(grid [][]int) int {
	m, n := len(grid), len(grid[0])
	dirs := []int{-1, 0, 1, 0, -1}

	var dfs func(i, j int)
	dfs = func(i, j int) {
		grid[i][j] = 2
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1 {
				dfs(x, y)
			}
		}
	}

	count := func() int {
		cnt := 0
		for i := 0; i < m; i++ {
			for j := 0; j < n; j++ {
				if grid[i][j] == 1 {
					dfs(i, j)
					cnt++
				}
			}
		}
		for i := 0; i < m; i++ {
			for j := 0; j < n; j++ {
				if grid[i][j] == 2 {
					grid[i][j] = 1
				}
			}
		}
		return cnt
	}

	if count() != 1 {
		return 0
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 {
				grid[i][j] = 0
				if count() != 1 {
					return 1
				}
				grid[i][j] = 1
			}
		}
	}
	return 2
}