Number of Islands II
<p>You are given an empty 2D binary grid <code>grid</code> of size <code>m x n</code>. The grid represents a map where <code>0</code>&#39;s represent water and <code>1</code>&#39;s represent land. Initially, all the cells of <code>grid</code> are water cells (i.e., all the cells are <code>0</code>&#39;s).</p>

<p>We may perform an add land operation which turns the water at position into a land. You are given an array <code>positions</code> where <code>positions[i] = [r<sub>i</sub>, c<sub>i</sub>]</code> is the position <code>(r<sub>i</sub>, c<sub>i</sub>)</code> at which we should operate the <code>i<sup>th</sup></code> operation.</p>

<p>Return <em>an array of integers</em> <code>answer</code> <em>where</em> <code>answer[i]</code> <em>is the number of islands after turning the cell</em> <code>(r<sub>i</sub>, c<sub>i</sub>)</code> <em>into a land</em>.</p>

<p>An <strong>island</strong> is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0300-0399/0305.Number%20of%20Islands%20II/images/tmp-grid.jpg" style="width: 500px; height: 294px;" />
<pre>
<strong>Input:</strong> m = 3, n = 3, positions = [[0,0],[0,1],[1,2],[2,1]]
<strong>Output:</strong> [1,1,2,3]
<strong>Explanation:</strong>
Initially, the 2d grid is filled with water.
- Operation #1: addLand(0, 0) turns the water at grid[0][0] into a land. We have 1 island.
- Operation #2: addLand(0, 1) turns the water at grid[0][1] into a land. We still have 1 island.
- Operation #3: addLand(1, 2) turns the water at grid[1][2] into a land. We have 2 islands.
- Operation #4: addLand(2, 1) turns the water at grid[2][1] into a land. We have 3 islands.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> m = 1, n = 1, positions = [[0,0]]
<strong>Output:</strong> [1]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m, n, positions.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>positions[i].length == 2</code></li>
	<li><code>0 &lt;= r<sub>i</sub> &lt; m</code></li>
	<li><code>0 &lt;= c<sub>i</sub> &lt; n</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you solve it in time complexity <code>O(k log(mn))</code>, where <code>k == positions.length</code>?</p>
Union Find; Array; Hash Table
class UnionFind {
    private final int[] p;
    private final int[] size;

    public UnionFind(int n) {
        p = new int[n];
        size = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            size[i] = 1;
        }
    }

    public int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }

    public boolean union(int a, int b) {
        int pa = find(a), pb = find(b);
        if (pa == pb) {
            return false;
        }
        if (size[pa] > size[pb]) {
            p[pb] = pa;
            size[pa] += size[pb];
        } else {
            p[pa] = pb;
            size[pb] += size[pa];
        }
        return true;
    }
}

class Solution {
    public List<Integer> numIslands2(int m, int n, int[][] positions) {
        int[][] grid = new int[m][n];
        UnionFind uf = new UnionFind(m * n);
        int[] dirs = {-1, 0, 1, 0, -1};
        int cnt = 0;
        List<Integer> ans = new ArrayList<>();
        for (var p : positions) {
            int i = p[0], j = p[1];
            if (grid[i][j] == 1) {
                ans.add(cnt);
                continue;
            }
            grid[i][j] = 1;
            ++cnt;
            for (int k = 0; k < 4; ++k) {
                int x = i + dirs[k], y = j + dirs[k + 1];
                if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == 1
                    && uf.union(i * n + j, x * n + y)) {
                    --cnt;
                }
            }
            ans.add(cnt);
        }
        return ans;
    }
}