<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 UnionFind:

    def __init__(self, n):
        self.tree = list(range(n))
        self.rank = [0]*n
    
    def find(self, i):
        if self.tree[i] != i:
            self.tree[i] = self.find(self.tree[i])
        return self.tree[i]
    
    def union(self, i, j):
        i, j = self.find(i), self.find(j)
        if i == j:
            return
        
        if self.rank[i] <= self.rank[j]:
            self.tree[i] = j
            self.rank[j] += self.rank[i]
        else:
            self.tree[j] = i
            self.rank[i] += self.rank[j]
    
    def get_components(self):
        components = defaultdict(list)
        for i in range(len(self.tree)):
            root = self.find(i)
            components[root].append(i)
        
        return components
    
class Solution:
    def groupStrings(self, words: List[str]) -> List[int]:

        def to_mask(word):
            mask = 0
            for char in word:
                mask |= (1 << (ord(char) - 97))
            
            return mask

        UF = UnionFind(len(words))
        word_dict = defaultdict(list)

        for i, word in enumerate(words):
            word_dict[to_mask(word)].append(i)
        
        for word in word_dict:
            for index in range(1, len(word_dict[word])):
                UF.union(word_dict[word][index], word_dict[word][index - 1])

        for i, word in enumerate(words):
            mask = to_mask(word)
            in_word, not_in_word = [], []
            for j in range(26):
                if (mask >> j) & 1:
                    in_word.append(j)
                else:
                    not_in_word.append(j)
            new_masks = []
            for j in in_word:
                new_masks.append(mask - (1 << j))
            for j in not_in_word:
                new_masks.append(mask + (1 << j))
            for j, k in itertools.product(in_word, not_in_word):
                new_masks.append(mask - (1 << j) + (1 << k))
            for new_mask in new_masks:
                if new_mask in word_dict:
                    UF.union(i, word_dict[new_mask][0])

        components = UF.get_components()

        return [len(components), max(len(group) for group in components.values())]
        

            



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