<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>

namespace UnionFind {
struct SetElement {
  int root;
  int size;
};

class DisjointSet {
 public:
  DisjointSet(int n) {
    for (int i = 0; i < n; i++)
      forest.emplace_back(i, 1);
  }

  int find(int e) {
    if (forest[e].root != e)
      forest[e].root = find(forest[e].root);

    return forest[e].root;
  }

  void setUnion(int e1, int e2) {
    int r1{find(e1)}, r2{find(e2)};

    if (r1 == r2)
      return;

    if (forest[r1].size < forest[r2].size)
      swap(r1, r2);

    forest[r2].root = r1;
    forest[r1].size += forest[r2].size;
  }

 private:
  vector<SetElement> forest;
};
}

using namespace UnionFind;

class Solution {
public:
  int numSimilarGroups(vector<string>& strs) {
    int n{static_cast<int>(strs.size())};
    int l{static_cast<int>(strs[0].length())};
    DisjointSet ds(n);

    if (l < n) {
      unordered_map<string, int> um;

      for (int i = 0; i < n; i++)
        um[strs[i]] = i;

      for (int i = 0; i < n; i++) {
        string s{strs[i]};
        ds.setUnion(i, um[s]);

        for (int j = 0; j < l; j++) {
          for (int k = j + 1; k < l; k++) {
            swap(s[j], s[k]);

            if (um.find(s) != um.end())
              ds.setUnion(i, um[s]);

            swap(s[j], s[k]);
          }
        }
      }
    } else {
      for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
          int flag{0};
          char character2Swap[2];

          for (int k = 0; k < l && flag < 3; k++) {
            if (strs[i][k] != strs[j][k]) {
              switch (flag) {
              case 0:
                character2Swap[0] = strs[i][k];
                character2Swap[1] = strs[j][k];
                flag = 1;
                break;

              case 1:
                flag = (strs[i][k] == character2Swap[1] && strs[j][k] == character2Swap[0])? 2 : 3;
                break;

              case 2:
                flag = 3; 
              }
            }
          }

          if (flag < 3)
            ds.setUnion(i, j);
        }
      }
    }

    vector<bool> isRoot(n, false);
    int output{0};

    for (int i = 0; i < n; i++) {
      if (!isRoot[ds.find(i)]) {
        isRoot[ds.find(i)] = true;
        output++;
      }
    }
    
    return output;
  }
};
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