<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>> dir = {{1, 0}, {0, -1}, {-1, 0}, {0, 1}};
    
    void dfs(int i, int j, vector<vector<int>>& grid, vector<vector<bool>>& visited, int val) {
        grid[i][j] = val;
        visited[i][j] = true;
        int n = grid.size();
        for (int k = 0; k < 4; k++) {
            int ni = i + dir[k][0];
            int nj = j + dir[k][1];
            if (ni >= 0 && ni < n && nj >= 0 && nj < n && grid[ni][nj] && !visited[ni][nj])
                dfs(ni, nj, grid, visited, val);
        }
    }

    int largestIsland(vector<vector<int>>& grid) {
        int islands = 0;
        int n = grid.size();
        vector<vector<bool>> visited(n, vector<bool>(n, false));

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] && !visited[i][j])
                    dfs(i, j, grid, visited, ++islands);
            }
        }

        vector<int> freq(islands + 1, 0);
        int result = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                freq[grid[i][j]]++;
                if (grid[i][j]) result = max(result, freq[grid[i][j]]);
            }
        }

        for (int i = 0 ; i < n; i++) {
            for (int j = 0; j < n; j++) {     
                if (grid[i][j] == 0) {
                    set<int> s;
                    for (int k = 0; k < 4; k++) {
                        int ni = i + dir[k][0];
                        int nj = j + dir[k][1];
                        if (ni >= 0 && ni <n && nj >= 0 && nj < n && grid[ni][nj])
                                s.insert(grid[ni][nj]);
                    }
                    int curr = 1;
                    for (auto it = s.begin(); it != s.end(); it++) curr += freq[*it];
                    result = max(result, curr);
                }
            }
        }
        return result;
    }
};
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