<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>> dp;
    
    // Depth-First Search to calculate the size of an island
    int dfs(int i, int j, vector<vector<int>>& grid, unordered_set<int>& set, unordered_set<int>& setg) {
        int n = grid.size();
        if (i < 0 || i >= n || j < 0 || j >= n || grid[i][j] == 0 || setg.find(i * n + j) != setg.end()) 
            return 0;
        
        if (dp[i][j] > 0) 
            return dp[i][j];
        
        set.insert(i * n + j);
        setg.insert(i * n + j);
        grid[i][j] = 0;
        
        vector<pair<int, int>> ng = {{i + 1, j}, {i, j + 1}, {i, j - 1}, {i - 1, j}};
        int curr = 1;
        
        for (auto [x, y] : ng) {
            curr += dfs(x, y, grid, set, setg);
        }
        
        grid[i][j] = 1;
        dp[i][j] = curr;
        return curr;
    }

    int largestIsland(vector<vector<int>>& grid) {
        int ans = 0;
        int n = grid.size();
        dp.resize(n, vector<int>(n, 0));
        vector<pair<int, int>> ng = {{1, 0}, {0, 1}, {0, -1}, {-1, 0}};
        unordered_map<int, int> islandSize; // To store size of islands
        int islandId = 2; // To identify different islands
        
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1 && dp[i][j] == 0) {
                    unordered_set<int> setg;
                    unordered_set<int> set;
                    int size = dfs(i, j, grid, set, setg);
                    islandSize[islandId] = size;
                    for (int k : set) {
                        dp[k / n][k % n] = islandId;
                    }
                    ++islandId;
                }
            }
        }

        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    unordered_set<int> neighborIslands;
                    int sum = 1; // Account for the current cell turning to 1
                    for (auto [dx, dy] : ng) {
                        int x = i + dx, y = j + dy;
                        if (x >= 0 && x < n && y >= 0 && y < n && dp[x][y] > 0) {
                            neighborIslands.insert(dp[x][y]);
                        }
                    }
                    for (int id : neighborIslands) {
                        sum += islandSize[id];
                    }
                    ans = max(ans, sum);
                }
            }
        }
        
        if (ans == 0) return n * n;
        return ans;
    }
};
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