<p>Given an <code>m x n</code> 2D binary grid <code>grid</code> which represents a map of <code>&#39;1&#39;</code>s (land) and <code>&#39;0&#39;</code>s (water), return <em>the number of islands</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>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;]
]
<strong>Output:</strong> 1
</pre>

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

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;]
]
<strong>Output:</strong> 3
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>grid[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class UnionFind:
    def __init__(self, grid):
        m, n = len(grid), len(grid[0])
        self.parent, self.rank = [], []
        self.count = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == "1":
                    self.count += 1
                    self.parent.append(i*n + j)
                else:
                    self.parent.append(-1)
                self.rank.append(1)

    def find(self, i):
        if self.parent[i] != i:
            self.parent[i] = self.find(self.parent[i])
        return self.parent[i]

    def union(self, i,j):
        ires, jres = self.find(i), self.find(j)
        if ires == jres:
            return
        
        irank, jrank = self.rank[ires], self.rank[jres]
        if irank < jrank:
            self.parent[ires] = jres
            self.rank[jrank] += 1
        else:
            self.parent[jres] = ires
            self.rank[irank] += 1
        self.count -= 1
        
    def getCount(self):
        return self.count


class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        m, n = len(grid), len(grid[0])
        UF = UnionFind(grid)
        directions = [(0,1), (0,-1), (1,0), (-1,0)]

        for i in range(m):
            for j in range(n):
                if grid[i][j] == "1":
                    grid[i][j] = "0"
                    for direction in directions:
                        ni, nj = i+direction[0], j+direction[1]
                        if ni < 0 or ni >=m or nj < 0 or nj >= n or grid[ni][nj] != "1":
                            continue
                        UF.union(i*n + j, ni*n + nj)

        return UF.getCount()

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