<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 dp[501][501];
    int dx[4] = {0, 0, -1, 1};
    int dy[4] = {1, -1, 0, 0};
    
    int dfs(int x, int y, int id, vector<vector<int>> &a) {
        int n = a.size();
        if (x < 0 || y < 0 || x >= n || y >= n || a[x][y] != 1 || dp[x][y] != 0) 
            return 0;
        dp[x][y] = id;
        int siz = 1;
        for (int i = 0; i < 4; i++) {
            siz += dfs(x + dx[i], y + dy[i], id, a);
        }
        return siz;
    }
    
    int largestIsland(vector<vector<int>>& a) {
        int n = a.size();
        memset(dp, 0, sizeof(dp));
        unordered_map<int, int> large;
        int cnt = 2;  // Start from 2 to avoid confusion with the initial values
        int ans = 0;

        // Step 1: Identify and label all islands
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (a[i][j] == 1 && dp[i][j] == 0) {
                    int siz = dfs(i, j, cnt, a);
                    large[cnt++] = siz;
                    ans = max(ans, siz);
                }
            }
        }

        // Step 2: Calculate potential largest island by flipping one '0' to '1'
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (a[i][j] == 0) {
                    unordered_set<int> connected;
                    int temp = 1;  // start with size 1 for the flipped '0'
                    for (int d = 0; d < 4; d++) {
                        int nx = i + dx[d], ny = j + dy[d];
                        if (nx >= 0 && ny >= 0 && nx < n && ny < n && dp[nx][ny] > 0) {
                            connected.insert(dp[nx][ny]);
                        }
                    }
                    for (int id : connected) {
                        temp += large[id];
                    }
                    ans = max(ans, temp);
                }
            }
        }
        
        return ans;
    }
};

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