<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 DisjointSet {
public:
    int n;
    int numComponents;
    vector<int> size, parent;
    
    DisjointSet(int n) {
        numComponents = n;
        size.resize(n+1, 1); // Initialize size to 1 for each element
        parent.resize(n+1);
        for(int i = 0; i <= n; i++) {
            parent[i] = i;
        }    
    }
    
    int findUPar(int a) {
        if(parent[a] == a) {
            return a;
        }
        return parent[a] = findUPar(parent[a]); // Path compression
    }
  
    void unionBySize(int u, int v) {
        int pu = findUPar(u);
        int pv = findUPar(v);
        if(pu != pv) { // Only union if they are in different sets
            if(size[pu] < size[pv]) {
                parent[pu] = pv;
                size[pv] += size[pu];
            } else {
                parent[pv] = pu;
                size[pu] += size[pv];
            }
            numComponents--;
        }
    }
};

class Solution {
public:
    bool isSimilar(string a, string b) {
        int n = a.size();
        int diffCount = 0;
        vector<int> diffIndices;
        
        for(int i = 0; i < n; i++) {
            if(a[i] != b[i]) {
                diffCount++;
                if(diffCount > 2) return false;  // More than two differences, not similar
                diffIndices.push_back(i);
            }
        }
        
        // Check if swapping the differing characters would make the strings equal
        if(diffCount == 2) {
            if(a[diffIndices[0]] == b[diffIndices[1]] && a[diffIndices[1]] == b[diffIndices[0]]) {
                return true;
            }
            return false;
        }
        
        return diffCount == 0;  // Strings with no differences are similar
    }
    
    int numSimilarGroups(vector<string>& strs) {
        int n = strs.size();
        DisjointSet ds(n);
        for(int i = 0; i < n; i++) {
            for(int j = i+1; j < n; j++) {
                if(isSimilar(strs[i], strs[j])) {
                    ds.unionBySize(i, j);
                }
            }
        }
        return ds.numComponents;
    }
};

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