Shortest Bridge
<p>You are given an <code>n x n</code> binary matrix <code>grid</code> where <code>1</code> represents land and <code>0</code> represents water.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>&#39;s not connected to any other <code>1</code>&#39;s. There are <strong>exactly two islands</strong> in <code>grid</code>.</p>

<p>You may change <code>0</code>&#39;s to <code>1</code>&#39;s to connect the two islands to form <strong>one island</strong>.</p>

<p>Return <em>the smallest number of </em><code>0</code><em>&#39;s you must flip to connect the two islands</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,1],[1,0]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [[0,1,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> grid = [[1,1,1,1,1],[1,0,0,0,1],[1,0,1,0,1],[1,0,0,0,1],[1,1,1,1,1]]
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length == grid[i].length</code></li>
	<li><code>2 &lt;= n &lt;= 100</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There are exactly two islands in <code>grid</code>.</li>
</ul>
Depth-First Search; Breadth-First Search; Array; Matrix
func shortestBridge(grid [][]int) (ans int) {
	n := len(grid)
	dirs := []int{-1, 0, 1, 0, -1}
	type pair struct{ i, j int }
	q := []pair{}
	var dfs func(int, int)
	dfs = func(i, j int) {
		grid[i][j] = 2
		q = append(q, pair{i, j})
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if x >= 0 && x < n && y >= 0 && y < n && grid[x][y] == 1 {
				dfs(x, y)
			}
		}
	}
	for i, x := 0, 1; i < n && x == 1; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 {
				dfs(i, j)
				x = 0
				break
			}
		}
	}
	for {
		for i := len(q); i > 0; i-- {
			p := q[0]
			q = q[1:]
			for k := 0; k < 4; k++ {
				x, y := p.i+dirs[k], p.j+dirs[k+1]
				if x >= 0 && x < n && y >= 0 && y < n {
					if grid[x][y] == 1 {
						return
					}
					if grid[x][y] == 0 {
						grid[x][y] = 2
						q = append(q, pair{x, y})
					}
				}
			}
		}
		ans++
	}
}