<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>

class Solution:
    def largestIsland(self, grid: List[List[int]]) -> int:
        def is_island(ij):
            i, j = ij
            return grid[i][j] == 1

        def neighbors(ij):
            i, j = ij
            if i > 0:
                yield (i - 1, j)
            if i < len(grid) - 1:
                yield (i + 1, j)
            if j > 0:
                yield (i, j - 1)
            if j < len(grid[0]) - 1:
                yield (i, j + 1)

        parents = {ij: ij for ij in product(range(len(grid)), range(len(grid[0]))) if is_island(ij)}
        def find(a):
            while a != parents[a]:
                parents[a] = parents[parents[a]]
                a = parents[a]
            return a

        sizes = {ij: 1 for ij in product(range(len(grid)), range(len(grid[0]))) if is_island(ij)}
        def union(a, b):
            pa, pb = find(a), find(b)
            if pa == pb:
                return
            if sizes[pa] > sizes[pb]:
                parents[pb] = pa
                sizes[pa] += sizes[pb]
            else:
                parents[pa] = pb
                sizes[pb] += sizes[pa]

        for a in product(range(len(grid)), range(len(grid[0]))):
            if is_island(a):
                for b in neighbors(a):
                    if is_island(b):
                        union(a, b)

        def combined_size(islands):
            return sum(sizes[pa] for pa in set(find(a) for a in islands))
        
        res = max(sizes.values()) if sizes else 0
        for a in product(range(len(grid)), range(len(grid[0]))):
            if not is_island(a):
                res = max(res, 1 + combined_size(b for b in neighbors(a) if is_island(b)))

        return res

52468