<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]:
        # words: each have unique letters
        # s1, s2 are connected if
        #    {s1} + c == {s2}
        #    {s1} - c == {s2}
        #    {s1'} == {s2} after swapping one letter for another

        # union-find with adjacency?

        # only care about sets, e.g. bit sets
        # can do union-find on bit sets
        # then count up the number of words with each bit set

        # add letter: set a new bit to true
        # remove a letter: toggle off a bit
        # replace letter: toggle off, toggle on another

        # for replacing a character:
        #    for each character we have
        #    and for each character we don't have
        #    that's a new combination
        #
        #    worst case: have 13 / 26
        #    so we'd try 169 combinations each
        # not too bad, annoying but by no means impossible

        counts = defaultdict(int)
        oa = ord('a')
        for word in words:
            b = 0
            for c in word:
                b |= 1 << (ord(c) - oa)
            counts[b] += 1

        parent = {b: -1 for b in counts}
        
        def find(b: int) -> int:
            if parent[b] < 0: return b
            parent[b] = find(parent[b])
            return parent[b]

        def union(b1: int, b2: int) -> None:
            b1 = find(b1)
            b2 = find(b2)

            if b1 == b2: return

            if parent[b2] < parent[b1]: b1, b2 = b2, b1

            parent[b1] += parent[b2]
            parent[b2] = b1

        ALL = (1 << 26) - 1

        def iter_bits(bits: int) -> Iterator[int]:
            while bits: 
                yield bits & -bits
                bits &= bits-1

        for b in counts:
            missing = ALL ^ b

            # we'll get add later when we find a longer string and subtract a letter
            # # for each missing character, add to set
            # for add in iter_bits(missing):
            #     b_add = b | add
            #     if b_add in counts: union(b, b_add)

            # for each present character: remove from set
            for rem in iter_bits(b):
                b_rem = b ^ rem
                if b_rem in counts: union(b, b_rem)

            for add in iter_bits(missing):
                for rem in iter_bits(b):
                    if rem < add: continue # instead of e.g "ab" -> "ac" and "ac"->"ab" we'll just do one
                    b_swap = b ^ add ^ rem
                    if b_swap in counts: union(b, b_swap)

        groups = defaultdict(int) # root -> number of words
        for b, f in counts.items():
            b = find(b)
            groups[b] += f

        return [len(groups), max(groups.values())]
32600