<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:
    vector<vector<int>> dirs = {{0,1}, {1,0}, {0,-1}, {-1,0}};

    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size(), group = 2; // Start group numbering from 2
        
        // Assign group numbers using DFS
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    dfs(grid, i, j, group);
                    group++;
                }
            }
        }

        // Calculate the size of each group
        vector<int> sizeg(group, 0);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] > 1) {
                    sizeg[grid[i][j]]++;
                }
            }
        }

        // Initialize answer with the largest island size without flipping
        int ans = 0;
        for (int i = 2; i < group; ++i) {
            ans = max(ans, sizeg[i]);
        }

        // Check the adjacent areas of the ungrouped (0) nodes
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    set<int> s; // Set to store unique adjacent group numbers
                    for (auto dir : dirs) {
                        int x = i + dir[0], y = j + dir[1];
                        if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] > 1) {
                            s.insert(grid[x][y]);
                        }
                    }
                    int t = 1; // Start with the flipped 0
                    for (int x : s) {
                        t += sizeg[x];
                    }
                    ans = max(ans, t);
                }
            }
        }
        return ans;
    }

    void dfs(vector<vector<int>>& grid, int x, int y, int group) {
        grid[x][y] = group;
        int n = grid.size();
        for (auto dir : dirs) {
            int i = x + dir[0], j = y + dir[1];
            if (i >= 0 && i < n && j >= 0 && j < n && grid[i][j] == 1) {
                dfs(grid, i, j, group);
            }
        }
    }
};

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