<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 {
public:
    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<int> area;
        vector<vector<int>> id(n, vector<int>(n, 0));
        int index = 2, maxArea = 0;
        
        // DFS to calculate the area of each island
        function<int(int, int)> dfs = [&](int x, int y) -> int {
            if (x < 0 || x >= n || y < 0 || y >= n || grid[x][y] != 1) return 0;
            grid[x][y] = index;
            int a = 1;
            vector<pair<int, int>> directions = {{0,1}, {1,0}, {0,-1}, {-1,0}};
            for (auto& dir : directions) {
                a += dfs(x + dir.first, y + dir.second);
            }
            return a;
        };
        
        // Mark each island with a unique index and calculate its area
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    area.push_back(dfs(i, j));
                    maxArea = max(maxArea, area.back());
                    ++index;
                }
            }
        }
        
        // Check the possibility of flipping a 0 to 1 and connecting islands
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    unordered_set<int> seen;
                    int possibleArea = 1;  // The flipped 0 to 1
                    vector<pair<int, int>> directions = {{0,1}, {1,0}, {0,-1}, {-1,0}};
                    for (auto& dir : directions) {
                        int x = i + dir.first, y = j + dir.second;
                        if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] > 1 && !seen.count(grid[x][y])) {
                            seen.insert(grid[x][y]);
                            possibleArea += area[grid[x][y] - 2];
                        }
                    }
                    maxArea = max(maxArea, possibleArea);
                }
            }
        }
        
        return maxArea;
    }
};

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