<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 {
    vector<int> parent;
    vector<int> rank;
    vector<int> size;
    public:
    DisjointSet(int n){
        // 0 based indexing
        parent.resize(n, 0);
        rank.resize(n, 0);
        size.resize(n,0);
        for(int i=0;i<n;i++){
            parent[i] = i;
            size[i] = 1;
        }
    }
    
    int findUpar(int node){
        if(parent[node] == node) return node;
        // path compression
        return parent[node] = findUpar(parent[node]);
    }
    
    void unionByRank(int u, int v){
        int p_u = findUpar(u);
        int p_v = findUpar(v);
        if(p_u == p_v) return;
        if(rank[p_u]>rank[p_v]){
            parent[p_v] = p_u;
        }else if(rank[p_u]<rank[p_v]){
            parent[p_u] = p_v;
        }else {
            parent[p_v] = p_u;
            rank[p_u]++;
        }
    }
    
    void unionBySize(int u, int v){
        int p_u = findUpar(u);
        int p_v = findUpar(v);
        if(p_u == p_v) return;
        if(size[p_u]>size[p_v]){
            parent[p_v] = p_u;
            size[p_u] += size[p_v];
        }else {
            parent[p_u] = p_v;
            size[p_v] += size[p_u];
        }
    }
};

class Solution {
public:
    bool helper(string s1, string s2){
        int count = 0;
        for(int i=0;i<min(s1.size(), s2.size());i++){
            if(s1[i] != s2[i])count++;
        }
        return count<=2;
    }
    int numSimilarGroups(vector<string>& strs) {
        DisjointSet ds(strs.size());
        int n= strs.size();
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                if(helper(strs[i], strs[j])){
                    ds.unionByRank(i,j);
                }
            }
        }

        int count = 0;
        for(int i=0;i<n;i++){
            if(ds.findUpar(i) == i){
                count++;
            }
        }
        return count;
    }
};
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