<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, A: List[List[int]]) -> int:
        n      = len(A)
        parent = [-1] * (n * n)
        rank   = [ 0] * (n * n)

        def id(i, j):
            return i * n + j

        def find(u):
            if u == parent[u]: return parent[u]
            parent[u] = find(parent[u])
            return parent[u]

        def union(u, v):
            pu, pv = find(u), find(v)
            if pu != pv:
                if rank[pu] >= rank[pv]:
                    parent[pv] = pu
                    rank[pu]  += rank[pv]
                else:
                    parent[pu] = pv
                    rank[pv]  += rank[pu]

        for i in range(n):
            for j in range(n):
                if A[i][j]: 
                    idx = id(i, j)
                    parent[idx], rank[idx] = idx, 1

        res = 1
        for i in range(n):
            for j in range(n):
                if A[i][j]:
                    idx = id(i, j)
                    if i > 0     and A[i - 1][j]: union(idx, id(i - 1, j))
                    if j > 0     and A[i][j - 1]: union(idx, id(i, j - 1))
                    if i < n - 1 and A[i + 1][j]: union(idx, id(i + 1, j))
                    if j < n - 1 and A[i][j + 1]: union(idx, id(i, j + 1))
                    res = max(res, rank[find(idx)])

        for i in range(n):
            for j in range(n):
                if not A[i][j]:
                    S = set()
                    if i > 0     and A[i - 1][j]: S.add(find(id(i - 1, j)))
                    if j > 0     and A[i][j - 1]: S.add(find(id(i, j - 1)))
                    if i < n - 1 and A[i + 1][j]: S.add(find(id(i + 1, j)))
                    if j < n - 1 and A[i][j + 1]: S.add(find(id(i, j + 1)))
                    if S: res = max(res, sum(rank[idx] for idx in S) + 1)
        return res

        
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