<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 dfs(vector<vector<int>>& grid, int i, int j, int id) {
        int n = grid.size();
        if (i < 0 || i >= n || j < 0 || j >= n || grid[i][j] != 1) {
            return 0;
        }
        grid[i][j] = id;  // Mark the cell with the current island id
        return 1 + dfs(grid, i + 1, j, id) + dfs(grid, i - 1, j, id) +
                   dfs(grid, i, j + 1, id) + dfs(grid, i, j - 1, id);
    }

    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        unordered_map<int, int> islandArea;  // Store area of each island with its id
        int island_id = 2;  // Start from 2 as island identifiers
        int max_area = 0;

        // First pass: Identify all islands and their areas
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    int area = dfs(grid, i, j, island_id);
                    islandArea[island_id] = area;
                    max_area = max(max_area, area);
                    island_id++;
                }
            }
        }

        // Second pass: Evaluate the largest island possible by flipping a 0 to 1
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    unordered_set<int> seen;  // To store unique neighboring island ids
                    int area = 1;  // Start with the area 1 (flipping 0 to 1)
                    for (auto [x, y] : vector<pair<int, int>>{{i+1, j}, {i-1, j}, {i, j+1}, {i, j-1}}) {
                        if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] > 1) {
                            seen.insert(grid[x][y]);
                        }
                    }
                    for (int id : seen) {
                        area += islandArea[id];
                    }
                    max_area = max(max_area, area);
                }
            }
        }

        return max_area;
    }
};
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