<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:
        areas = defaultdict(int)
        ROWS, COLS  = len(grid), len(grid[0])
        
        DIRECTIONS = [(1, 0), (0, 1), (-1, 0), (0, -1)]
        
        # (*) to track adjesent to the coastline cells
        adj = defaultdict(set)
    
        def dfs(r, c, island_num):
            if r not in range(0, ROWS) or c not in range(0, COLS) or grid[r][c] != 1:
                return
            grid[r][c] = island_num
            areas[island_num] += 1
            for d in DIRECTIONS:
                new_r, new_c = r + d[0], c + d[1]
                dfs(new_r, new_c, island_num)
                # (*)
                if new_r in range(0, ROWS) and new_c in range(0, COLS) and grid[new_r][new_c] == 0:
                    adj[island_num].add((new_r, new_c))

        tmp_id = 2
        for r in range(0, ROWS):
            for c in range(0, COLS):
                if grid[r][c] == 1:
                    dfs(r, c, tmp_id)
                    tmp_id += 1

        if not areas:
            return 1

        res = 0
        for i in range(2, tmp_id + 1):
            island_area = areas[i]
            cells = adj[i]
            best_cell_area = island_area
            for cell in cells:
                connected = set()
                extra_area = 1 # from 0 -> 1 flip
                for d in DIRECTIONS:
                    new_r, new_c = cell[0] + d[0], cell[1] + d[1]
                    if (new_r in range(0, ROWS) and new_c in range(0, COLS) 
                        and grid[new_r][new_c] != 0 and grid[new_r][new_c] != i and grid[new_r][new_c] not in connected):
                        #candidate cell connects with some other island
                        extra_area += areas[grid[new_r][new_c]]
                        connected.add(grid[new_r][new_c])
                best_cell_area = max(best_cell_area, island_area + extra_area)
                
            res = max(res, best_cell_area)
        return res

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