<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 m, n;
    vector<int> dx = {1, -1, 0, 0};
    vector<int> dy = {0, 0, 1, -1};
    void dfs(int i, int j, vector<vector<int>> &grid, int ind, int &ct){
        grid[i][j] = ind;
        ct++;
        
        for(int k = 0; k < 4; k++){
            int nx = i + dx[k];
            int ny = j + dy[k];
            if(nx >= 0 && ny >= 0 && nx < m && ny < n && grid[nx][ny] == 1){
                dfs(nx, ny, grid, ind, ct);
            }
        }
    }
    int largestIsland(vector<vector<int>>& grid) {
        m = grid.size();
        n = grid[0].size();
        int ind = 2;
        unordered_map<int, int> mp;

        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    int ct = 0;
                    dfs(i, j, grid, ind, ct);
                    mp[ind] = ct;
                    ind++;
                }
            }
        }


        queue<pair<int, int>> q;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 0){
                    q.push({i, j});
                }
            }
        }

        if(q.size() == 0){
            return m*n;
        }
        
        int mx = -1;
        while(!q.empty()){
            auto [i, j] = q.front();
            q.pop();
            set<int> st;
            int curr = 1;
            for(int k = 0; k < 4; k++){
                int nx = i + dx[k];
                int ny = j + dy[k];
                if(nx >= 0 && ny >= 0 && nx < m && ny < n && grid[nx][ny] != 0){
                    if(st.find(grid[nx][ny]) != st.end()){
                        continue;
                    }
                    curr += mp[grid[nx][ny]];
                    st.insert(grid[nx][ny]);
                }
            }

            mx = max(mx, curr);
        }

        return mx;
    }
};
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