<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>

#include <vector>
#include <algorithm>

using namespace std;

class DisjointSet {
    vector<int> rank, parent, size;

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

    int findpar(int node) {
        if (node != parent[node]) {
            parent[node] = findpar(parent[node]); // Path compression
        }
        return parent[node];
    }

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

        if (ulp_u != ulp_v) {
            if (size[ulp_u] > size[ulp_v]) {
                parent[ulp_v] = ulp_u;
                size[ulp_u] += size[ulp_v];
            } else {
                parent[ulp_u] = ulp_v;
                size[ulp_v] += size[ulp_u];
            }
        }
    }

    int getSize(int u) {
        return size[findpar(u)];
    }
};

class Solution {
public:
    int largestIsland(vector<vector<int>>& grid) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);
        int rows = grid.size();
        int cols = grid[0].size();
        int n = rows * cols;
        
        // Check if the grid is all ones
        bool allOnes = true;
        for (int i = 0; i < rows && allOnes; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 0) {
                    allOnes = false;
                    break;
                }
            }
        }
        if (allOnes) {
            return n; // Return the total number of cells as all cells are part of the island
        }

        DisjointSet ds(n);
        vector<pair<int, int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        // Union find on initial grid
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 1) {
                    for (auto direc : directions) {
                        int ni = i + direc.first;
                        int nj = j + direc.second;
                        if (ni >= 0 && ni < rows && nj >= 0 && nj < cols && grid[ni][nj] == 1) {
                            ds.unionBySize(i * cols + j, ni * cols + nj);
                        }
                    }
                }
            }
        }

        int total = 0;

        // Check all zeros and compute the possible size of the island if turned to 1
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 0) {
                    int current = 1; // Start with the current cell itself
                    vector<int> uniqueParents;
                    for (auto direc : directions) {
                        int ni = i + direc.first;
                        int nj = j + direc.second;
                        if (ni >= 0 && ni < rows && nj >= 0 && nj < cols && grid[ni][nj] == 1) {
                            int root = ds.findpar(ni * cols + nj);
                            if (find(uniqueParents.begin(), uniqueParents.end(), root) == uniqueParents.end()) {
                                uniqueParents.push_back(root);
                                current += ds.getSize(ni * cols + nj);
                            }
                        }
                    }
                    total = max(total, current);
                }
            }
        }

        return total;
    }
};

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