<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:
        m, n = len(grid), len(grid[0])

        # bfs to group idx
        groups = []
        visited = set()

        def bfs(i, j):
            q = deque([(i, j)])
            visited.add((i, j))
            cur_visited = []
            while q:
                i, j = q.popleft()
                cur_visited.append((i, j))
                for _dir in [[1,0], [-1,0], [0,1], [0,-1]]:
                    n_i, n_j = i + _dir[0], j + _dir[1]
                    if 0 <= n_i < m and 0 <= n_j < n and grid[n_i][n_j] == 1 and (n_i, n_j) not in visited:
                        visited.add((n_i, n_j))
                        q.append((n_i, n_j))
            return cur_visited

        for i in range(m):
            for j in range(n):
                if (i, j) not in visited and grid[i][j] == 1:
                    groups.append(bfs(i, j))

        # find border for each group
        def border(group):
            res = []
            for i, j in group:
                for _dir in [[1,0], [-1,0], [0,1], [0,-1]]:
                    n_i, n_j = i + _dir[0], j + _dir[1]
                    if 0 <= n_i < m and 0 <= n_j < n and grid[n_i][n_j] == 0:
                        res.append((i, j))
                        break
            return res
        borders = [border(group) for group in groups]

        # each border idx send its island size to its neighbor 0
        # sum island sizes for each neighbor 0, return the max
        group_sizes = [len(group) for group in groups]
        neighbor2size = {}

        def send_size(border, group_size, visited):
            for i, j in border:
                for _dir in [[1,0], [-1,0], [0,1], [0,-1]]:
                    n_i, n_j = i + _dir[0], j + _dir[1]
                    if 0 <= n_i < m and 0 <= n_j < n and grid[n_i][n_j] == 0 and (n_i, n_j) not in visited:
                        # no repeated send
                        visited.add((n_i, n_j))
                        neighbor2size[(n_i, n_j)] = neighbor2size.get((n_i, n_j), 0) + group_size

        for border, group_size in zip(borders, group_sizes):
            send_size(border, group_size, set())

        # find max
        if len(neighbor2size) == 0:
            if grid[0][0] == 0: return 1
            else: return m * n
        else: return max(list(neighbor2size.values())) + 1
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