<p>You are given a <strong>0-indexed</strong> array of strings <code>words</code>. Each string consists of <strong>lowercase English letters</strong> only. No letter occurs more than once in any string of <code>words</code>.</p>

<p>Two strings <code>s1</code> and <code>s2</code> are said to be <strong>connected</strong> if the set of letters of <code>s2</code> can be obtained from the set of letters of <code>s1</code> by any <strong>one</strong> of the following operations:</p>

<ul>
	<li>Adding exactly one letter to the set of the letters of <code>s1</code>.</li>
	<li>Deleting exactly one letter from the set of the letters of <code>s1</code>.</li>
	<li>Replacing exactly one letter from the set of the letters of <code>s1</code> with any letter, <strong>including</strong> itself.</li>
</ul>

<p>The array <code>words</code> can be divided into one or more non-intersecting <strong>groups</strong>. A string belongs to a group if any <strong>one</strong> of the following is true:</p>

<ul>
	<li>It is connected to <strong>at least one</strong> other string of the group.</li>
	<li>It is the <strong>only</strong> string present in the group.</li>
</ul>

<p>Note that the strings in <code>words</code> should be grouped in such a manner that a string belonging to a group cannot be connected to a string present in any other group. It can be proved that such an arrangement is always unique.</p>

<p>Return <em>an array</em> <code>ans</code> <em>of size</em> <code>2</code> <em>where:</em></p>

<ul>
	<li><code>ans[0]</code> <em>is the <strong>maximum number</strong> of groups</em> <code>words</code> <em>can be divided into, and</em></li>
	<li><code>ans[1]</code> <em>is the <strong>size of the largest</strong> group</em>.</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;b&quot;,&quot;ab&quot;,&quot;cde&quot;]
<strong>Output:</strong> [2,3]
<strong>Explanation:</strong>
- words[0] can be used to obtain words[1] (by replacing &#39;a&#39; with &#39;b&#39;), and words[2] (by adding &#39;b&#39;). So words[0] is connected to words[1] and words[2].
- words[1] can be used to obtain words[0] (by replacing &#39;b&#39; with &#39;a&#39;), and words[2] (by adding &#39;a&#39;). So words[1] is connected to words[0] and words[2].
- words[2] can be used to obtain words[0] (by deleting &#39;b&#39;), and words[1] (by deleting &#39;a&#39;). So words[2] is connected to words[0] and words[1].
- words[3] is not connected to any string in words.
Thus, words can be divided into 2 groups [&quot;a&quot;,&quot;b&quot;,&quot;ab&quot;] and [&quot;cde&quot;]. The size of the largest group is 3.  
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;ab&quot;,&quot;abc&quot;]
<strong>Output:</strong> [1,3]
<strong>Explanation:</strong>
- words[0] is connected to words[1].
- words[1] is connected to words[0] and words[2].
- words[2] is connected to words[1].
Since all strings are connected to each other, they should be grouped together.
Thus, the size of the largest group is 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 26</code></li>
	<li><code>words[i]</code> consists of lowercase English letters only.</li>
	<li>No letter occurs more than once in <code>words[i]</code>.</li>
</ul>

class DS:
  def __init__(self, n):
    self.p = [i for i in range(n)]
    self.r = [0 for _ in range(n)]
  
  def find(self, x):
    if self.p[x] != x:
      self.p[x] = self.find(self.p[x])
    return self.p[x]
  
  def union(self, x, y):
    x, y = self.find(x), self.find(y)
    if x == y: return
    rx, ry = self.r[x], self.r[y]
    if rx > ry:
      self.p[y] = x
    elif ry > rx:
      self.p[x] = y
    else:
      self.p[x] = y
      self.r[y] += 1

@cache
def get_index(ch):
  return ord(ch)-ord('a')

@cache
def get_mask(s):
  mask = 0
  for ch in s:
    mask |= 1 << get_index(ch)
  return mask

class Solution:
  def get_neighbors(self, s, word_masks):
    mask = get_mask(s)
    res = {word_masks[mask]} 
    # add one letter
    for i in range(26):
      if mask & (1 << i) == 0:
        new_mask = mask | (1 << i)
        if new_mask in word_masks:
          res.add(word_masks[new_mask])
    # delete one letter
    for i in range(26):
      if mask & (1 << i) != 0:
        new_mask = mask ^ (1 << i)
        if new_mask in word_masks:
          res.add(word_masks[new_mask])
    # replace one letter
    for i in range(26):
      if mask & (1 << i) != 0:
        for j in range(26):
          new_mask = mask ^ (1 << i)
          new_mask |= 1 << j
          if new_mask in word_masks:
            res.add(word_masks[new_mask])
    return res

  def groupStrings(self, words: List[str]) -> List[int]:
    n = len(words)
    uf = DS(n)
    word_masks = dict()
    for i, w in enumerate(words):
      word_masks[get_mask(w)] = i
    # print(word_masks)
    for i, w in enumerate(words):
      for nei in self.get_neighbors(w, word_masks):
        uf.union(i, nei)
    for i in range(n):
      uf.find(i)
    # print(uf.p)
    c = Counter(uf.p)
    # print(c)
    return [len(c), max(c.values())] 
        
42600