<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> rank, parent, size;
    DisjointSet(int n){
        rank.resize(n+1,0);
        parent.resize(n+1,0);
        size.resize(n+1,0);
        for(int i=0;i<=n;i++){
            parent[i]=i;
            size[i]=1;
        }
    }
    int findPar(int node){
        if(node==parent[node])  return node;
        return parent[node] = findPar(parent[node]);
    }
    void UnionByRank(int u, int v){
        int ulp_u = findPar(u);
        int ulp_v = findPar(v);
        if(ulp_v==ulp_u) return;
        if(rank[ulp_v]>rank[ulp_u]) {
            parent[ulp_u] = ulp_v;
        }else if(rank[ulp_v]<rank[ulp_u]) {
            parent[ulp_v] = ulp_u;
        }else{
            parent[ulp_v] = ulp_u;
            rank[ulp_u]++;
        }
    }
    void UnionBySize(int u, int v){
        int ulp_u = findPar(u);
        int ulp_v = findPar(v);
        if(ulp_v==ulp_u) return;
        if(size[ulp_v]>size[ulp_u]) {
            parent[ulp_u] = ulp_v;
            size[ulp_v] += size[ulp_u];
        }else{
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v];
        }
    }
};

class Solution {
public:
    bool isValid(int i, int j, int n){
        return i>=0 && j>=0 && i<n && j<n;
    }
    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        DisjointSet ds(n*n);
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==0) continue;
                vector<int> di = {-1,0,1,0};
                vector<int> dj = {0,1,0,-1};
                for(int ind=0;ind<4;ind++){
                    int ni = i+di[ind];
                    int nj = j+dj[ind];
                    if(isValid(ni,nj,n) && grid[ni][nj]==1) {
                        int  nodeNo = (i*n)+j;
                        int adjNo = (ni*n)+nj;
                        ds.UnionBySize(nodeNo,adjNo);
                    }
                }
            }
        }
        int mx = 0;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1) continue;
                set<int>comp;
                vector<int> di = {-1,0,1,0};
                vector<int> dj = {0,1,0,-1};
                for(int ind=0;ind<4;ind++){
                    int ni = i+di[ind];
                    int nj = j+dj[ind];
                    if(isValid(ni,nj,n)) {
                        if(grid[ni][nj]==1){
                            comp.insert(ds.findPar(ni*n+nj));
                        }
                    }
                }
                int sizeTot = 0;
                for(auto it:comp){
                    sizeTot+=ds.size[it];
                }
                mx=max(mx,sizeTot+1);
            }
        }
        for (int cellNo = 0; cellNo < n * n; cellNo++) {
            mx = max(mx, ds.size[ds.findPar(cellNo)]);
        }
        return mx;
    }
};
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