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

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

    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];
        }
    }
};

class Solution {
public:
    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        DisjointSet ds(n * n);
        int drow[] = {-1, 0, 1, 0};
        int dcol[] = {0, 1, 0, -1};
        int maxSize = 0;

        // Step 1: Union all adjacent '1's in the grid
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    int node = i * n + j;
                    for (int k = 0; k < 4; k++) {
                        int nrow = i + drow[k];
                        int ncol = j + dcol[k];
                        if (nrow >= 0 && ncol >= 0 && nrow < n && ncol < n && grid[nrow][ncol] == 1) {
                            int adjNode = nrow * n + ncol;
                            ds.unionBySize(node, adjNode);
                        }
                    }
                }
            }
        }

        // Step 2: Evaluate the maximum island size considering flipping a '0' to '1'
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0) {
                    int node = i * n + j;
                    int sizeSet = 1; // size of the flipped '0'
                    int seen[4] = {-1, -1, -1, -1};  // Track 4 possible unique components

                    for (int k = 0; k < 4; k++) {
                        int nrow = i + drow[k];
                        int ncol = j + dcol[k];
                        if (nrow >= 0 && ncol >= 0 && nrow < n && ncol < n && grid[nrow][ncol] == 1) {
                            int adjNode = nrow * n + ncol;
                            int par = ds.findUPar(adjNode);

                            // Avoid double counting
                            bool found = false;
                            for (int m = 0; m < 4; m++) {
                                if (seen[m] == par) {
                                    found = true;
                                    break;
                                }
                            }
                            if (!found) {
                                sizeSet += ds.size[par];
                                seen[k] = par;
                            }
                        }
                    }
                    maxSize = max(maxSize, sizeSet);
                }
            }
        }

        // If grid is fully '1', check the largest component
        if (maxSize == 0) {
            for (int cellNo = 0; cellNo < n * n; cellNo++) {
                maxSize = max(maxSize, ds.size[ds.findUPar(cellNo)]);
            }
        }

        return maxSize;
    }
};

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