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:
    def __init__(self, n: int):
        self.p = list(range(n))
        self.size = [1] * n

    def find(self, x: int):
        if self.p[x] != x:
            self.p[x] = self.find(self.p[x])
        return self.p[x]

    def union(self, a: int, b: int) -> bool:
        pa, pb = self.find(a - 1), self.find(b - 1)
        if pa == pb:
            return False
        if self.size[pa] > self.size[pb]:
            self.p[pb] = pa
            self.size[pa] += self.size[pb]
        else:
            self.p[pa] = pb
            self.size[pb] += self.size[pa]
        return True


class Solution:
    def numIslands2(self, m: int, n: int, positions: List[List[int]]) -> List[int]:
        uf = UnionFind(m * n)
        grid = [[0] * n for _ in range(m)]
        ans = []
        dirs = (-1, 0, 1, 0, -1)
        cnt = 0
        for i, j in positions:
            if grid[i][j]:
                ans.append(cnt)
                continue
            grid[i][j] = 1
            cnt += 1
            for a, b in pairwise(dirs):
                x, y = i + a, j + b
                if (
                    0 <= x < m
                    and 0 <= y < n
                    and grid[x][y]
                    and uf.union(i * n + j, x * n + y)
                ):
                    cnt -= 1
            ans.append(cnt)
        return ans
