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

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

    void unionByRank(int u, int v) {
        int ulParentV = findParent(v);
        int ulParentU = findParent(u);

        if (ulParentU == ulParentV) return;
        if (rank[ulParentU] < rank[ulParentV]) {
            parent[ulParentU] = ulParentV;
        } else if (rank[ulParentU] > rank[ulParentV]) {
            parent[ulParentV] = ulParentU;
        } else {
            parent[ulParentV] = ulParentU;
            rank[ulParentU]++;
        }
        cmp--;
    }
    int part(){
        return cmp;
    }
};

class Solution {
public:

    void bfs(int i , vector<vector<int>> & adj , vector<int> & visit){
        queue<int> q;
        q.push(i);
        visit[i] = true;

        while(!q.empty()){
            int node = q.front();q.pop();

            for(auto i : adj[node]){
                if(!visit[i]){
                    q.push(i);
                    visit[i]=true;
                }
            }
        }
    }

    bool similar(string & a , string & b){
        int len = a.size(),cnt=0;
        for(int i=0;i<len;i++){
            if(a[i]!=b[i])  cnt++;
        }
        return cnt<=2;
    }
    int numSimilarGroups(vector<string>& str) {

        vector<vector<int>> adj(str.size());

        int len = str.size();

        for(int i=0;i<len;i++){
            for(int j=i+1;j<len;j++){
                if(similar(str[i],str[j])){
                    adj[i].push_back(j);
                    adj[j].push_back(i);
                }
            }
        }

        vector<int> visit(str.size());

        // dsu ds(str.size());
        // for(int i=0;i<str.size();i++){
        //     for(auto j : adj[i]){
        //         ds.unionByRank(i,j);
        //     }
        // }

        // return ds.part();

        int cnt = 0;
        for(int i=0;i<len;i++){
            if(!visit[i]){
                bfs(i,adj,visit);
                cnt++;
            }
        }
        return cnt;
    }
};
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