<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>> directions = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    
    int dfs(int x, int y, vector<vector<int>>& grid, vector<vector<int>>& islandMap, int islandId) {
        int n = grid.size();
        int size = 1;
        islandMap[x][y] = islandId;  // Mark this cell as part of the island
        for (auto& dir : directions) {
            int nx = x + dir[0];
            int ny = y + dir[1];
            if (nx >= 0 && ny >= 0 && nx < n && ny < n && grid[nx][ny] == 1 && islandMap[nx][ny] == 0) {
                size += dfs(nx, ny, grid, islandMap, islandId);
            }
        }
        return size;
    }

    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> islandMap(n, vector<int>(n, 0)); // Map to store island IDs
        unordered_map<int, int> islandSize;  // Store the size of each island
        int islandId = 2; // Start labeling islands from 2 because 0 and 1 are already used in the grid

        // Step 1: Find all islands and label them
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1 && islandMap[i][j] == 0) {
                    int size = dfs(i, j, grid, islandMap, islandId);
                    islandSize[islandId] = size;
                    ++islandId;
                }
            }
        }

        // Step 2: Check for the largest island if no change is made
        int maxIsland = 0;
        for (const auto& [id, size] : islandSize) {
            maxIsland = max(maxIsland, size);
        }

        // Step 3: Try converting each '0' to '1' and calculate the new island size
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    unordered_set<int> neighborIslands;
                    int newSize = 1;  // This '0' becomes '1', so start with size 1
                    for (auto& dir : directions) {
                        int nx = i + dir[0];
                        int ny = j + dir[1];
                        if (nx >= 0 && ny >= 0 && nx < n && ny < n && islandMap[nx][ny] > 0) {
                            neighborIslands.insert(islandMap[nx][ny]);
                        }
                    }
                    // Sum up sizes of all distinct neighboring islands
                    for (int islandId : neighborIslands) {
                        newSize += islandSize[islandId];
                    }
                    maxIsland = max(maxIsland, newSize);
                }
            }
        }

        // If the entire grid was already filled with 1s
        return maxIsland == 0 ? n * n : maxIsland;
    }
};
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