Making A Large Island
<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>
Depth-First Search; Breadth-First Search; Union Find; Array; Matrix
class Solution {
public:
    int largestIsland(vector<vector<int>>& grid) {
        int n = grid.size();
        int p[n][n];
        memset(p, 0, sizeof(p));
        int cnt[n * n + 1];
        memset(cnt, 0, sizeof(cnt));
        const int dirs[5] = {-1, 0, 1, 0, -1};
        int root = 0;
        int ans = 0;
        function<int(int, int)> dfs = [&](int i, int j) {
            p[i][j] = root;
            ++cnt[root];
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k];
                int y = j + dirs[k + 1];
                if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] && !p[x][y]) {
                    dfs(x, y);
                }
            }
            return cnt[root];
        };
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] && !p[i][j]) {
                    ++root;
                    ans = max(ans, dfs(i, j));
                }
            }
        }
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                if (!grid[i][j]) {
                    unordered_set<int> s;
                    for (int k = 0; k < 4; ++k) {
                        int x = i + dirs[k];
                        int y = j + dirs[k + 1];
                        if (x >= 0 && x < n && y >= 0 && y < n) {
                            s.insert(p[x][y]);
                        }
                    }
                    int t = 1;
                    for (int x : s) {
                        t += cnt[x];
                    }
                    ans = max(ans, t);
                }
            }
        }
        return ans;
    }
};