<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 {
public:
    vector<int> par,size;
    int find(int a){
        if(par[a]==a)
        return a;
        return par[a]=find(par[a]);
    }
    void unon(int i, int j){
        int par_i=find(i);
        int par_j=find(j);

        if(par_i!=par_j){
            if(size[par_i]>size[par_j]){
                par[par_j]=par[par_i];
                size[par_i]+=size[par_j];
            }
            else{
                par[par_i]=par[par_j];
                size[par_j]+=size[par_i];
            }
        }
    }
    bool is_con(string &a, string &b){
        int n=a.size();
        int count=0;
        vector<int> diff;
        for(int i=0;i<n;i++){
            if(a[i]!=b[i]){
                count++;
                if(count>=3)
                return false;
                diff.push_back(i);
            }
            
        }
        if(count==0)
        return true;
        if(count==2&&(a[diff[0]]==b[diff[1]]&&a[diff[1]]==b[diff[0]]))
        return true;
        return false;
    }
    int numSimilarGroups(vector<string>& strs) {
        int n=strs.size();
        size.assign(n,1);
        par.resize(n);
        for(int i=0;i<n;i++){
            par[i]=i;
        }
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                if(is_con(strs[i],strs[j])){
                    unon(i,j);
                }
            }
        }
        
        unordered_set<int> st;
        for(int i=0;i<n;i++){
            st.insert(find(i));
        }
        return st.size();

    }
};
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