<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 Node {
public:
    Node* Links[26];
    int end;
    Node(){
        end = -1;
        for(int i=0;i<26;++i)
            Links[i] = nullptr;
    }
};

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

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

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

vector<int> inGroup;
vector<int> group;
int n;
void match(int i, pair<int,int>& val, string& s, Node* root){
    if(i==n){
        if(val.first == -1 || val.first == -2){
            inGroup.push_back(root->end);
        }
        return;
    }
    if(root->Links[s[i]-'a']){
        match(i+1, val, s, root->Links[s[i]-'a']);
    }

    if(val.first == -2){
        return;
    }
    else{
        if(val.first==-1){
            for(int itr=0;itr<26;++itr){
                if(s[i]-'a' != itr && root->Links[itr]){
                    val = {itr, s[i]-'a'};
                    match(i+1, val, s, root->Links[itr]);
                    val = {-1, -1};
                }
            }
        }
        else{
            if(auto nxt = root->Links[val.second]; s[i]-'a' == val.first && nxt){
                val = {-2, -2};
                match(i+1, val, s, nxt);
            }
        }
    }
}

class Solution {
public:
    int numSimilarGroups(vector<string>& strs) {
        n = strs[0].size();
        DisjointSet ds(strs.size());
        Node* root = new Node();
        pair<int,int> p;
        for(int i=0;i<strs.size();++i){
            inGroup.resize(0);
            int end = -1;
            p = {-1,-1};
            match(0, p, strs[i], root);
            if(inGroup.size()==0){
                end = group.size()+1;
            }
            else{
                cout<<i+1<<": ";
                for(auto idx: inGroup){
                    cout<<idx<<" ";
                    ds.unionByRank(i+1, idx);
                }
                cout<<"\n";
                end = ds.findUPar(i+1);
            }

            auto tmp = root;

            for(auto c: strs[i]){
                if(tmp->Links[c-'a']==nullptr){
                    tmp->Links[c-'a'] = new Node();
                }
                tmp= tmp->Links[c-'a'];
            }
            tmp->end = i+1;
        }

        set<int> st;
        for(int i=1;i<=strs.size();++i)
            st.insert(ds.findUPar(i));
        return st.size();
    }
};
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