<p>Two strings, <code>X</code> and <code>Y</code>, are considered similar if either they are identical or we can make them equivalent by swapping at most two letters (in distinct positions) within the string <code>X</code>.</p>

<p>For example, <code>&quot;tars&quot;</code>&nbsp;and <code>&quot;rats&quot;</code>&nbsp;are similar (swapping at positions <code>0</code> and <code>2</code>), and <code>&quot;rats&quot;</code> and <code>&quot;arts&quot;</code> are similar, but <code>&quot;star&quot;</code> is not similar to <code>&quot;tars&quot;</code>, <code>&quot;rats&quot;</code>, or <code>&quot;arts&quot;</code>.</p>

<p>Together, these form two connected groups by similarity: <code>{&quot;tars&quot;, &quot;rats&quot;, &quot;arts&quot;}</code> and <code>{&quot;star&quot;}</code>.&nbsp; Notice that <code>&quot;tars&quot;</code> and <code>&quot;arts&quot;</code> are in the same group even though they are not similar.&nbsp; Formally, each group is such that a word is in the group if and only if it is similar to at least one other word in the group.</p>

<p>We are given a list <code>strs</code> of strings where every string in <code>strs</code> is an anagram of every other string in <code>strs</code>. How many groups are there?</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> strs = [&quot;tars&quot;,&quot;rats&quot;,&quot;arts&quot;,&quot;star&quot;]
<strong>Output:</strong> 2
</pre>

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

<pre>
<strong>Input:</strong> strs = [&quot;omv&quot;,&quot;ovm&quot;]
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= strs.length &lt;= 300</code></li>
	<li><code>1 &lt;= strs[i].length &lt;= 300</code></li>
	<li><code>strs[i]</code> consists of lowercase letters only.</li>
	<li>All words in <code>strs</code> have the same length and are anagrams of each other.</li>
</ul>

class Solution:
    def numSimilarGroups(self, strs: List[str]) -> int:
        def isIdentical(s1, s2):
            count = 0
            for i in range(len(s1)):
                if s1[i] != s2[i]:
                    count += 1

                if count > 2:
                    return False

            return True

        n = len(strs)
        uf = UnionFind(n)
        for i in range(n):
            for j in range(i + 1, n):
                if isIdentical(strs[i], strs[j]):
                    uf.union(i, j)

        return uf.components


class UnionFind:
    def __init__(self, size):
        self.root = [i for i in range(size)]
        self.rank = [0]*size
        self.components = size

    def find(self, x):
        while x != self.root[x]:
            x = self.root[x]

        return x

    def union(self, x, y):
        rootX = self.find(x)
        rootY = self.find(y)
        if rootX == rootY:
            return

        if self.rank[rootX] < self.rank[rootY]:
            self.root[rootX] = rootY
        elif self.rank[rootY] < self.rank[rootX]:
            self.root[rootY] = rootX
        else:
            self.root[rootY] = rootX
            self.rank[rootX] += 1

        self.components -= 1
16800