<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:
    vector<int> rank, parent, size;
    DisjointSet(int n) {
        rank.resize(n+1, 0); 
        parent.resize(n+1);
        size.resize(n+1); 
        for(int i = 0;i<=n;i++) {
            parent[i] = i; 
            size[i] = 1; 
        }
    }

    int findUPar(int node) {
        if(node == parent[node])
            return node; 
        return parent[node] = findUPar(parent[node]); 
    }

    void unionBySize(int u, int v) {
        int ulp_u = findUPar(u); 
        int ulp_v = findUPar(v); 
        if(ulp_u == ulp_v) return; 
        if(size[ulp_u] < size[ulp_v]) {
            parent[ulp_u] = ulp_v; 
            size[ulp_v] += size[ulp_u]; 
        }
        else {
            parent[ulp_v] = ulp_u;
            size[ulp_u] += size[ulp_v]; 
        }
    }
};

class Solution {
public:

    bool check(string a,string b)
    { int n=a.length();
        int diff=0;

        for(int i=0;i<n;i++)
        {
            if(a[i]-b[i]!=0) diff++;
        }

        return diff==2 || diff==0;
    }
    


    int numSimilarGroups(vector<string>& strs) {
        // X,Y -> two strings
        // we will apply DSU here

        // for loop chlaakar check kar lo ki kon konse connect ho skte hai and then we will 
        // connect them using DSU. now see in the description of the Question it is given that
        // atleast ek word se match krna chahiye toh shi hai na 
        // if u see a DSU tars-rats and rats-arts so therefore we will see a groups tars-rats-arts
        //SIMPLE DSU LOGIC -> solve it

        int n=strs.size();
        DisjointSet ds(n);

        for(int i=0;i<n-1;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                bool result=check(strs[i],strs[j]);
                if(result)
                {
                    ds.unionBySize(i,j);
                }
            }
        }
        int ans=0;
        for(int i=0;i<n;i++)
        {
            if(ds.parent[i]==i) ans++;
        }

        return ans;
    }
};
108418