<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{
private:
    vector<int> par;vector<int> size;
public:
    DisjointSet(int n){
        size.resize(n,1);
        for(int i=0;i<n;i++){
            par.push_back(i);
        }
    }
    int findUparent(int a){
         if(par[a]==a) return a;
         return par[a]=findUparent(par[a]);
    }

    void Union(int a,int b){
        int par1=findUparent(a);
        int par2=findUparent(b);
        if(par1==par2) return;
        if(size[par1]<size[par2]){
            par[par1]=par2;
            size[par2]+=size[par1];
        }
        else{
            par[par2]=par1;
            size[par1]+=size[par2];
        }
    }

    int grpno(){
        int ans=0;
        for(int i=0;i<par.size();i++){
            if(par[i]==i) ans++;
        }
        return ans;
    }




};
class Solution {
public:
    bool check(string a,string b){
    
        int cnt=0;
        for(int i=0;i<a.size();i++){
            if(a[i]!=b[i]) cnt++;
        }
        return cnt==2 || cnt==0;
    }
    int numSimilarGroups(vector<string>& strs) {
        int n=strs.size();
        DisjointSet ds(n);
        
         for(int i=0;i<strs.size();i++){
            for(int j=0;j<strs.size();j++){
                
                if(check(strs[i],strs[j])) {
                   ds.Union(i,j);
                }
                
            }
         }
         
         return ds.grpno();
        
    }
};
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