<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 parent[250001];
    int sz[250001];

    vector<pair<int, int>> mvmt{{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
    bool isV(int a, int b, int n) { return a >= 0 && b >= 0 && a < n && b < n; }

    void make(int v) {
        parent[v] = v;
        sz[v] = 1;
    }
    int find(int v) {
        if (parent[v] == v)
            return v;
        return parent[v] = find(parent[v]);
    }
    void Union(int a, int b) {
        a = find(a);
        b = find(b);
        if (a != b) {
            if (sz[a] < sz[b])
                swap(a, b);
            parent[b] = a;
            sz[a] += sz[b];
        }
    }
    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        // vector<vector<int>> sz(n, vector<int>(n, 0));
        int ans = 1;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                make(n * i + j);
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    for (auto mvt : mvmt) {
                        int x = i + mvt.first;
                        int y = j + mvt.second;
                        if (!isV(x, y, n))
                            continue;
                        if(grid[x][y]==0)   continue;
                        if(find(i*n+j)==find(x*n+y))    continue;
                        Union(i * n + j, x * n + y);
                    }
                    int r = find(i * n + j);
                    // cout<<r<<endl;
                    // cout<<sz[r]<<endl;
                    ans = max(ans, sz[r]);
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    vector<int> v;
                    int curr = 1;
                    for (auto mvt : mvmt) {
                        int x = i + mvt.first;
                        int y = j + mvt.second;
                        if (!isV(x, y, n))
                            continue;
                        if (grid[x][y] == 0)
                            continue;
                        // if(find(i*n+j)==find(x*n+y))    continue;
                        int z = find(x * n + y), cnter = 0;
                        for (auto it : v) {
                            if (it == z) {
                                cnter++;
                                break;
                            }
                        }
                        if (cnter == 0) {
                            curr += sz[z];
                        }
                        v.push_back(z);
                    }
                    ans = max(ans, curr);
                }
            }
        }
        return ans;
    }
};
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