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
class Solution:
    def largestIsland(self, grid: List[List[int]]) -> int:
        def dfs(i: int, j: int):
            p[i][j] = root
            cnt[root] += 1
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if 0 <= x < n and 0 <= y < n and grid[x][y] and p[x][y] == 0:
                    dfs(x, y)

        n = len(grid)
        cnt = Counter()
        p = [[0] * n for _ in range(n)]
        dirs = (-1, 0, 1, 0, -1)
        root = 0
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                if x and p[i][j] == 0:
                    root += 1
                    dfs(i, j)
        ans = max(cnt.values() or [0])
        for i, row in enumerate(grid):
            for j, x in enumerate(row):
                if x == 0:
                    s = set()
                    for a, b in pairwise(dirs):
                        x, y = i + a, j + b
                        if 0 <= x < n and 0 <= y < n:
                            s.add(p[x][y])
                    ans = max(ans, sum(cnt[root] for root in s) + 1)
        return ans
