<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 DisjointSets:
    def __init__(self):
        self.parents = []
        self.counts = []
        self.key_to_set = {}

    def add(self, key):
        if (s := self.key_to_set.get(key)) is None:
            s = len(self.parents)
            self.key_to_set[key] = s
            self.parents.append(s)
            self.counts.append(1)
        else:
            root = self._get_root(s)
            self.counts[root] += 1

    def merge(self, keys):
        parents = {self.key_to_set.get(key) for key in keys}
        parents.discard(None)
        roots = {self._get_root(p) for p in parents}
        root = max(roots, key=self.counts.__getitem__)
        self.counts[root] = sum(self.counts[r] for r in roots)
        for r in roots:
            self.parents[r] = root

    def _get_root(self, s):
        while self.parents[s] != s:
            s, self.parents[s] = self.parents[s], self.parents[self.parents[s]]
        return s

    def get_counts(self):
        counts = []
        for s, p in enumerate(self.parents):
            if s == p:
                counts.append(self.counts[s])
        return counts


class Solution:
    def groupStrings(self, words: list[str]) -> tuple[int, int]:
        groups = DisjointSets()

        for word in words:
            mask = 0
            for c in word:
                mask |= 1 << (ord(c) - ord('a'))
            groups.add(mask)

            keys = [mask]
            for i in range(26):
                 keys.append(mask ^ (1 << i))
                 if mask & (1 << i):
                    m = mask & ~(1<<i)
                    for j in range(26):
                        if not mask & (1 << j):
                            keys.append(m | (1 << j))

            groups.merge(keys)

        counts = groups.get_counts()
        return len(counts), max(counts)


def main():
    words = ["a","b","ab","cde"]
    print(Solution().groupStrings(words))


if __name__ == '__main__':
    main()

23500