<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 Solution:
    def groupStrings(self, words: List[str]) -> List[int]:
        from collections import defaultdict
        
        # Function to generate the bitmask for a given word
        def get_bitmask(word):
            mask = 0
            for char in word:
                mask |= 1 << (ord(char) - ord('a'))
            return mask
        
        # Union-Find data structure
        class UnionFind:
            def __init__(self, size):
                self.root = list(range(size))
                self.rank = [1] * size
                
            def find(self, x):
                if x == self.root[x]:
                    return x
                self.root[x] = self.find(self.root[x])
                return self.root[x]
                
            def union(self, x, y):
                rootX = self.find(x)
                rootY = self.find(y)
                if rootX != rootY:
                    if self.rank[rootX] > self.rank[rootY]:
                        self.root[rootY] = rootX
                    elif self.rank[rootX] < self.rank[rootY]:
                        self.root[rootX] = rootY
                    else:
                        self.root[rootY] = rootX
                        self.rank[rootX] += 1
                    
        # Step 1: Convert words to bitmasks and count them
        bitmasks = defaultdict(int)
        for word in words:
            bitmask = get_bitmask(word)
            bitmasks[bitmask] += 1
        
        # Step 2: Union the bitmasks based on connectivity rules
        uf = UnionFind(len(bitmasks))
        bitmask_to_id = {bitmask: i for i, bitmask in enumerate(bitmasks)}
        id_to_bitmask = {i: bitmask for i, bitmask in enumerate(bitmasks)}
        
        for bitmask, count in bitmasks.items():
            for i in range(26):
                # Adding or removing a letter
                new_bitmask = bitmask ^ (1 << i)
                if new_bitmask in bitmask_to_id:
                    uf.union(bitmask_to_id[bitmask], bitmask_to_id[new_bitmask])
                
                # Replacing a letter
                for j in range(i+1, 26):
                    if (bitmask & (1 << i)) and (bitmask & (1 << j)) == 0:
                        new_bitmask = bitmask ^ (1 << i) | (1 << j)
                        if new_bitmask in bitmask_to_id:
                            uf.union(bitmask_to_id[bitmask], bitmask_to_id[new_bitmask])
        
        # Step 3: Count the number of groups and the size of the largest group
        group_sizes = defaultdict(int)
        for bitmask in bitmasks:
            group_id = uf.find(bitmask_to_id[bitmask])
            group_sizes[group_id] += bitmasks[bitmask]
        
        num_groups = len(group_sizes)
        max_group_size = max(group_sizes.values())
        
        return [num_groups, max_group_size]

# Example usage:
solution = Solution()
words = ["a","b","ab","cde"]
print(solution.groupStrings(words))  # Output: [2, 3]

words = ["a","ab","abc"]
print(solution.groupStrings(words))  # Output: [1, 3]
24600