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
function largestIsland(grid: number[][]): number {
    const n = grid.length;
    const p = Array.from({ length: n }, () => Array(n).fill(0));
    const cnt = Array(n * n + 1).fill(0);
    const dirs = [-1, 0, 1, 0, -1];
    let root = 0;
    let ans = 0;

    const dfs = (i: number, j: number): number => {
        p[i][j] = root;
        cnt[root]++;
        for (let k = 0; k < 4; ++k) {
            const x = i + dirs[k];
            const y = j + dirs[k + 1];
            if (x >= 0 && x < n && y >= 0 && y < n && grid[x][y] === 1 && p[x][y] === 0) {
                dfs(x, y);
            }
        }
        return cnt[root];
    };

    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] === 1 && p[i][j] === 0) {
                root++;
                ans = Math.max(ans, dfs(i, j));
            }
        }
    }

    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < n; ++j) {
            if (grid[i][j] === 0) {
                const s = new Set<number>();
                for (let k = 0; k < 4; ++k) {
                    const x = i + dirs[k];
                    const y = j + dirs[k + 1];
                    if (x >= 0 && x < n && y >= 0 && y < n) {
                        s.add(p[x][y]);
                    }
                }
                let t = 1;
                for (const x of s) {
                    t += cnt[x];
                }
                ans = Math.max(ans, t);
            }
        }
    }
    return ans;
}
