<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{
    vector<int> parent, rank, size;
public:
    DisjointSet(int n){
        parent.resize(n+1);
        rank.resize(n+1, 0);
        size.resize(n+1);
        for(int i=0; i<=n; i++){
            parent[i] = i;
            size[i] = 1;
        }
    }

    int findUPar(int node){
        if(parent[node]==node) return node;
        return parent[node] = findUPar(parent[node]);
    }

    void unionByRank(int u, int v){
        int ulp_u = findUPar(u);
        int ulp_v = findUPar(v);

        if(ulp_u == ulp_v)return; // u and v are part of the same component

        if(rank[ulp_u] < rank[ulp_v]){
            parent[ulp_u] = ulp_v;
        }
        else if(rank[ulp_u] > rank[ulp_v]){
            parent[ulp_v] = ulp_u;
        }
        else{
            parent[ulp_v] = ulp_u;
            rank[ulp_u]++;
        }
    }

    void unionBySize(int u, int v){
        int ulp_u = findUPar(u);
        int ulp_v = findUPar(v);

        if(ulp_u == ulp_v)return;
        
        if(size[ulp_u] < size[ulp_v]){
            parent[ulp_u] = ulp_v;
            size[ulp_v] += size[ulp_u];
        }
        else{
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v];
        }
    }

    int sz(int node){
        return size[node];
    }
};
class Solution {
public:
    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        int idx = 0;
        vector<vector<int>>index(n, vector<int>(n, 0));
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                index[i][j] = idx;
                idx++;
            }
        }
        DisjointSet ds(idx);
        vector<int>di={-1, 0, 1, 0};
        vector<int>dj={0, 1, 0, -1};
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j]){
                    for(int k=0; k<4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(ni>=0&&ni<n&&nj>=0&&nj<n&&grid[ni][nj]){
                            ds.unionBySize(index[i][j], index[ni][nj]);
                        }
                    }
                }
            }
        }
        int res = 0;
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j]==0){
                    int ans = 0;
                    unordered_set<int>st;
                    for(int k=0; k<4; k++){
                        int ni = i+di[k];
                        int nj = j+dj[k];
                        if(ni>=0&&ni<n&&nj>=0&&nj<n&&grid[ni][nj]){
                            st.insert(ds.findUPar(index[ni][nj]));
                            // ds.unionBySize(index[i][j], index[ni][nj]);
                        }
                    }
                    for(auto it:st){
                        ans += ds.sz(it);
                    }
                    res = max(res, ans+1);
                }
            }
        }
        return (res==0)? n*n: res;
    }
};
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