<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 find(vector<int> &par, int x)
{
    return par[x]=par[x]==x?x:find(par,par[x]);
}

void Union(vector<int> &par, vector<int> &rank, int a, int b)
{
    a = find(par,a);
    b = find(par,b);

    if(rank[a]<rank[b])
    {
        rank[b]++;
        par[a]=b;
    }
    else
    {
        rank[a]++;
        par[b]=a;
    }
}
    int largestIsland(vector<vector<int>>& grid) {
        
        int n = grid.size();

        // vector<vector<bool>> vis(n,vector<bool>(n,false));

        vector<int> dx  = {1,-1,0,0};
        vector<int> dy = {0,0,1,-1};

        queue<pair<int,int>> qz;
        vector<int> par(n*n), rank(n*n);

        for(int i=0;i<n*n;i++)
        par[i]=i;

        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]==1)
                {
                    grid[i][j]=-1;
                    queue<pair<int,int>> q;
                    q.push({i,j});

                    while(q.size())
                    {
                        auto tp = q.front();
                        q.pop();
                        int x = tp.first;
                        int y = tp.second;

                        for(int d=0;d<4;d++)
                        {
                            int nr = x + dx[d];
                            int nc =  y + dy[d];

                            if(nr>=0 and nc>=0 and nr<n and nc<n and grid[nr][nc]==1)
                            {
                                q.push({nr,nc});
                                grid[nr][nc]=-1;
                                Union(par,rank,x*n+y,nr*n+nc);
                            }
                        }
                    }
                }
                if(grid[i][j]==0)
                qz.push({i,j});
            }
        }

        unordered_map<int,int> mp;
          int res = 0;
        for(int i=0;i<n*n;i++)
        {
            mp[find(par,i)]++;
            res = max(res, mp[find(par,i)]);
        }

        while(qz.size())
        {
            auto tp = qz.front();
            qz.pop();

            int x = tp.first;
            int y = tp.second;
            int ans = 0; 
            unordered_set<int> s;
            for(int i=0;i<4;i++)
            {
                int nr = x + dx[i];
                int nc = y + dy[i];

                if(nr>=0 and nc>=0 and nr<n and nc<n)
                {
                    int node = x*n+y;
                    int adjNode = nr*n+nc;
                    if(grid[nr][nc]==-1)
                    {
                        if(s.count(find(par,adjNode))==0)
                        {
                        ans += mp[find(par,adjNode)];
                        s.insert(find(par,adjNode));
                        }
                    }
                }
            }

            res = max (res, 1+ans);

        }

        return res;
    }
};
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