<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:
    vector<int> parent, size;
    
    // Find function with path compression
    int find(int x) {
        if (x != parent[x]) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }
    
    // Union function with union by size
    void unionSet(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        if (rootX != rootY) {
            if (size[rootX] >= size[rootY]) {
                parent[rootY] = rootX;
                size[rootX] += size[rootY];
            } else {
                parent[rootX] = rootY;
                size[rootY] += size[rootX];
            }
        }
    }
    
    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        parent.resize(n * n);
        size.resize(n * n, 1); // Each island initially has size 1 (for each '1')
        
        // Initialize the parent of each cell to itself
        for (int i = 0; i < n * n; ++i) {
            parent[i] = i;
        }
        
        // Step 1: Union adjacent '1's
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    int index = i * n + j;
                    
                    // Union with the right cell
                    if (j + 1 < n && grid[i][j + 1] == 1) {
                        unionSet(index, i * n + (j + 1));
                    }
                    
                    // Union with the down cell
                    if (i + 1 < n && grid[i + 1][j] == 1) {
                        unionSet(index, (i + 1) * n + j);
                    }
                }
            }
        }
        
        // Step 2: Find the largest island without any change
        int maxIslandSize = 0;
        unordered_map<int, int> islandSize;
        
        for (int i = 0; i < n * n; ++i) {
            if (grid[i / n][i % n] == 1) {
                int root = find(i);
                islandSize[root] = size[root];
                maxIslandSize = max(maxIslandSize, size[root]);
            }
        }
        
        // Step 3: Try flipping each '0' to '1' and calculate the resulting island size
        vector<int> dir = {-1, 0, 1, 0, -1}; // To traverse 4 directions
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 0) {
                    unordered_set<int> neighboringIslands;
                    int newIslandSize = 1; // The flipped '0' becomes part of the new island
                    
                    // Check all 4 neighbors
                    for (int d = 0; d < 4; ++d) {
                        int ni = i + dir[d], nj = j + dir[d + 1];
                        if (ni >= 0 && ni < n && nj >= 0 && nj < n && grid[ni][nj] == 1) {
                            int root = find(ni * n + nj);
                            neighboringIslands.insert(root);
                        }
                    }
                    
                    // Sum up the size of all distinct neighboring islands
                    for (int root : neighboringIslands) {
                        newIslandSize += islandSize[root];
                    }
                    
                    // Update maxIslandSize
                    maxIslandSize = max(maxIslandSize, newIslandSize);
                }
            }
        }
        
        return maxIslandSize == 0 ? n * n : maxIslandSize;
    }
};

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