<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 UnionFind:
    def __init__(self, size):
        self.parent = list(range(size))
        self.size = [1] * size
        self.components = size

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

        self.parent[index] = self.find(self.parent[index])
        return self.parent[index]

    def union(self, node1, node2) -> bool:
        root1 = self.find(node1)
        root2 = self.find(node2)

        if root1 != root2:
            self.components -= 1

            if self.size[root1] >  self.size[root2]:
                self.parent[root1] = root2
                self.size[root2] += self.size[root1]
            else:
                self.parent[root2] = root1
                self.size[root1] += self.size[root2]

        return root1 != root2

class Solution:
    def numSimilarGroups(self, strs: List[str]) -> int:
        n = len(strs)
        m = len(strs[0])

        uf = UnionFind(n)

        for i in range(n):
            for j in range(n):
                if i != j and self.identical(strs[i], strs[j], m):
                    uf.union(i, j)
        
        return uf.components

    def identical(self, word1: str, word2: str, size: int) -> bool:
        x, y = -1, -1
        for i in range(size):
            if word1[i] != word2[i]:
                if x == -1:
                    x = i       # first mismatch
                elif y == -1:
                    y = i       # second mismatch
                else:       
                    return False    # third mismatch
        # swapping should make them identical
        return word1[x] == word2[y] and word1[y] == word2[x]
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