<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:        
        # redo        
        areas = []
        surrounds = []
        
        def dfs(r, c, R, C, area, cover):
            area[0] += 1
            seen.add((r, c))
            for row, col in [(r-1, c), (r+1, c), (r, c-1), (r, c+1)]:
                if 0 <= row < R and 0 <= col < C:
                    if grid[row][col] and (row, col) not in seen:
                        dfs(row, col, R, C, area, cover)
                    elif grid[row][col] == 0:
                        cover.add((row, col))

        R, C = len(grid), len(grid[0])
        seen = set([])
        for r in range(R):
            for c in range(C):
                if (r, c) not in seen and grid[r][c]:
                    area = [0]
                    cover = set([])
                    dfs(r, c, R, C, area, cover)
                    areas.append(area[0])
                    surrounds.append(cover)

        max_area = 1
        for i, s in enumerate(areas):
            max_area = max(max_area, s + int(len(surrounds[i]) >= 1))

        counts = collections.defaultdict(int)
        for i, cover in enumerate(surrounds):
            for r, c in cover:
                counts[(r, c)] += areas[i]    
        max_area = max(max_area, 1 + max(counts.values() or [0]))  
        return max_area
        
        # Approach #2: Component Sizes
        # Time Complexity: O(N^2), where NN is the length and width of the grid.
        # Space Complexity: O(N^2), the additional space used in the depth first search by area.
        
        areas = []
        surrounds = [] # list of sets
        
        def dfs(r, c, R, C, area, cover):
            area[0] += 1
            seen.add((r, c))
            for row, col in [(r-1, c), (r+1, c), (r, c-1), (r, c+1)]:                 
                if 0 <= row < R and 0 <= col < C:
                    if grid[row][col] and (row, col) not in seen:
                        dfs(row, col, R, C, area, cover)
                    elif grid[row][col] == 0:
                        cover.add((row, col))            
            
        
        R, C = len(grid), len(grid[0])
        seen = set([])
        for r in range(R):
            for c in range(C):
                if (r, c) not in seen and grid[r][c]:
                    area = [0]
                    cover = set([]) 
                    dfs(r, c, R, C, area, cover)
                    areas.append(area[0])
                    surrounds.append(cover)
        
        maxArea = 1
        # add won't help the connection
        for i, s in enumerate(areas):
            maxArea = max(maxArea, s + int(len(surrounds[i]) >= 1) )
        
        countDict = collections.defaultdict(int)
        for i, cover in enumerate(surrounds):
            for r, c in cover:
                countDict[(r, c)] += areas[i]   
    
        return  max(maxArea, 1 + max(countDict.values() or [0]))                                   
                    

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