<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 dfs(vector<vector<int>>& grid, int r, int c, int index) {
        int n = grid.size();
        grid[r][c] = index;
        int size = 1;
        vector<int> dirR = {-1, 1, 0, 0};
        vector<int> dirC = {0, 0, -1, 1};

        for (int i = 0; i < 4; i++) {
            int newR = r + dirR[i];
            int newC = c + dirC[i];

            if (newR >= 0 && newR < n && newC >= 0 && newC < n && grid[newR][newC] == 1) {
                size += dfs(grid, newR, newC, index);
            }
        }

        return size;
    }

    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        unordered_map<int, int> island_size;
        int index = 2;

        // Step 1: label all islands with unique index and calculate their sizes
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    int size = dfs(grid, i, j, index);
                    island_size[index] = size;
                    index++;
                }
            }
        }

        // Step 2: Find the maximum island by size flipping one 0
        int max_size = 0;
        for (auto &[index, size] : island_size) {
            max_size = max(max_size, size);
        }

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0) {
                    unordered_set<int> seen;
                    int curr_size = 1;
                    vector<int> dirR = {-1, 1, 0, 0};
                    vector<int> dirC = {0, 0, -1, 1};

                    for (int k = 0; k < 4; k++) {
                        int newR = i + dirR[k];
                        int newC = j + dirC[k];
                        if (newR >= 0 && newR < n && newC >= 0 && newC < n && grid[newR][newC] > 1) {
                            seen.insert(grid[newR][newC]);
                        }
                    }

                    for (int idx: seen) {
                        curr_size += island_size[idx];
                    }

                    max_size = max(max_size, curr_size);
                }
            }
        }

        return max_size;
    }
};
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