<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 DisjointSet{
public:
    vector<int> size, par;
    DisjointSet(int n){
        size.resize(n+1, 1);
        par.resize(n+1);
        for(int i=0; i<par.size(); i++){
            par[i]=i;
        }
    }
    int findUPar(int a){
        if(a==par[a]){
            return a;
        }
        return par[a]=findUPar(par[a]);
    }
    void unionBySize(int a, int b){
        int par_a=findUPar(a);
        int par_b=findUPar(b);
        if(par_a == par_b){
            return;
        }
        if(size[par_a] < size[par_b]){
            par[par_a]=par_b;
            size[par_b] += size[par_a];
        }else{
            par[par_b]=par_a;
            size[par_a] += size[par_b];
        }
    }
    int findMax(){
        int maxi=INT_MIN;
        for(int i=0; i<size.size(); i++){
            maxi=max(maxi, size[i]);
        }
        return maxi;
    }
};
class Solution {
public:
    int largestIsland(vector<vector<int>>& grid) {
        int n=grid.size();
        vector<vector<int>> vis(n, vector<int>(n, 0));
        
        DisjointSet ds(n*n);
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j] == 1){
                    vector<pair<int, int>> dir={{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
                    for(auto it: dir){
                        int newx=i+it.first;
                        int newy=j+it.second;
                        if(newx >= 0 && newx < n && newy >=0 && newy < n && grid[newx][newy]==1){
                            int a= n * i + j;
                            int b= n * newx + newy;
                            ds.unionBySize(a, b);
                        }
                    }
                }
            }
        }
        int res=ds.findMax();
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j]==0){
                    unordered_set<int> st;
                    int ans=1;
                    vector<pair<int, int>> dir={{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
                    for(auto it: dir){
                        int newx=i+it.first;
                        int newy=j+it.second;
                        if(newx >= 0 && newx < n && newy >=0 && newy < n && grid[newx][newy]==1){
                            int a= n * newx + newy;
                            int par=ds.findUPar(a);
                            st.insert(par);
                        }
                    }
                    for(int elem: st){
                        ans=ans+ds.size[elem];
                    }
                    res=max(res, ans);
                }
            }
        }
        return res;
    }
};
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