<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:
        if not grid or not grid[0]:
            return 0
        m, n = len(grid), len(grid[0])
        directions = [(1, 0), (-1, 0), (0, 1), (0, -1)]
        cell2color = collections.defaultdict(int)
        color_count = collections.defaultdict(int)
        color = 2
        def dfs(r, c, color):
            if r < 0 or r >= m or c < 0 or c >= n or (r, c) in cell2color or grid[r][c] != 1:
                return
            cell2color[(r, c)] = color
            color_count[color] += 1
            for x, y in directions:
                nr, nc = x + r, y + c
                dfs(nr, nc, color)
        
        for r in range(m):
            for c in range(n):
                if grid[r][c] not in cell2color:
                    dfs(r, c, color)
                    color += 1
        
        has_zero = False
        res = 0
        for r in range(m):
            for c in range(n):
                if grid[r][c] == 0:
                    has_zero = True
                    neis = set()
                    curr = 1
                    for x, y in directions:
                        nr, nc = x + r, y + c
                        if nr < 0 or nr >= m or nc < 0 or nc >= n or grid[nr][nc] == 0:
                            continue
                        neis.add(cell2color[(nr, nc)])
                    for nei in neis:
                        curr += color_count[nei]
                    res = max(res, curr)
        return m * n if not has_zero else res
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