Making A Large Island
<p>You are given an <code>n x n</code> binary matrix <code>grid</code>. You are allowed to change <strong>at most one</strong> <code>0</code> to be <code>1</code>.</p>

<p>Return <em>the size of the largest <strong>island</strong> in</em> <code>grid</code> <em>after applying this operation</em>.</p>

<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>s.</p>

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

<pre>
<strong>Input:</strong> grid = [[1,0],[0,1]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1],[1,0]]
<strong>Output:</strong> 4
<strong>Explanation: </strong>Change the 0 to 1 and make the island bigger, only one island with area = 4.</pre>

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

<pre>
<strong>Input:</strong> grid = [[1,1],[1,1]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> Can&#39;t change any 0 to 1, only one island with area = 4.
</pre>

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

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n &lt;= 500</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; Union Find; Array; Matrix
func largestIsland(grid [][]int) int {
	n := len(grid)
	p := make([][]int, n)
	for i := range p {
		p[i] = make([]int, n)
	}
	cnt := make([]int, n*n+1)
	dirs := []int{-1, 0, 1, 0, -1}
	root := 0
	ans := 0

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

	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 1 && p[i][j] == 0 {
				root++
				ans = max(ans, dfs(i, j))
			}
		}
	}

	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			if grid[i][j] == 0 {
				s := make(map[int]struct{})
				for k := 0; k < 4; k++ {
					x := i + dirs[k]
					y := j + dirs[k+1]
					if x >= 0 && x < n && y >= 0 && y < n {
						s[p[x][y]] = struct{}{}
					}
				}
				t := 1
				for x := range s {
					t += cnt[x]
				}
				ans = max(ans, t)
			}
		}
	}
	return ans
}