<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 {
private:
vector<int> parent, size;

void start(int n){
    parent.resize(n);
    size.resize(n,1);
    for(int i = 0; i<n; i++){
        parent[i] = i;
    }
}

int findUpar(int x){
    if(parent[x] == x) return x;
    return parent[x] = findUpar(parent[x]);
}

void unionXY(int ux, int uy){

    if(size[ux] >= size[uy]){
        size[ux] += size[uy];
        parent[uy] = ux;
    }
    else{
        size[uy] += size[ux];
        parent[ux] = uy;
    }
}
public:
    bool isValid(int newr, int newc, int n) {
        return newr >= 0 && newr < n && newc >= 0 && newc < n;
    }

    int largestIsland(vector<vector<int>>& grid) {
        vector<int> dx = {0,0,-1,1};
        vector<int> dy = {-1,1,0,0};

        int n = grid.size();
        start(n*n);

        vector<vector<int>> zeros;
        for(int i = 0; i<n; i++){
            for(int j = 0; j<n; j++){
                if(grid[i][j]==0){
                    zeros.push_back({i,j});
                    continue;
                }
                for(int k = 0; k<4; k++){
                    int nx = i + dx[k], ny = j + dy[k];
                    if(isValid(nx,ny,n) && grid[nx][ny]==1){
                        int up1 = findUpar(i*n + j);
                        int up2 = findUpar(nx*n + ny);
                        if(up1!=up2){
                            unionXY(up1,up2);
                        }
                    }
                }
            }
        }

        if(zeros.size()==0){
            int maxS = INT_MIN;
            for(int i = 0; i<n*n; i++){
                if(parent[i]==i){
                    maxS = max(maxS, size[i]);
                }
            }
            return maxS;
        }

        int maxS = 0;
        for(auto i: zeros){
            int x = i[0], y = i[1];
            set<int> comp;
            for(int j = 0; j<4; j++){
                int nx = x + dx[j], ny = y + dy[j];
                if(isValid(nx,ny,n) && grid[nx][ny]==1){
                    comp.insert(findUpar(nx* n + ny));
                }
            }
            int sizeTotal = 0;
            for (auto it : comp) {
                sizeTotal += size[it];
            }

            maxS = max(maxS,sizeTotal+1);
        }
        return maxS;
    }
};
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