<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 DSU:
    def __init__(self, n: int):
        self.parent = list(range(n))
        self.rank = [1] * n
        self.size = [1] * n

    def find(self, x: int) -> int:
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]
    
    def union(self, x: int, y: int) -> bool:
        px, py = self.find(x), self.find(y)
        if px == py:
            return False
        elif self.rank[px] > self.rank[py]:
            self.parent[py] = px
            self.size[px] += self.size[py]
        elif self.rank[py] > self.rank[px]:
            self.parent[px] = py
            self.size[py] += self.size[px]
        else:
            self.parent[py] = px
            self.rank[px] += 1
            self.size[px] += self.size[py]
    
        return True

class Solution:
    def sort_string(self, s: str) -> str:
        return ''.join(sorted(s))

    def bit_mask(self, s: str) -> str:
        res = 0
        for c in s:
           res += 1 << (ord(c) - ord('a'))
        return res

    def groupStrings(self, words: List[str]) -> List[int]:
        n = len(words)
        dsu = DSU(n)

        words = [self.sort_string(word) for word in words]
        words = sorted(words, key = lambda x : (len(x), x))
        bit_masks = [self.bit_mask(word) for word in words]
        

        bitmask_2_idx = {}
        count = [0] * 27
        for i, (word, bit_mask) in enumerate(zip(words, bit_masks)):
            if bit_mask in bitmask_2_idx:
                dsu.union(bitmask_2_idx[bit_mask], i)
                continue

            # deleting one leter
            word_len = len(word)
            for j in range(word_len):
                # deleting:
                bit_loc = ord(word[j]) - ord('a')
                delete_bit_mask = bit_mask ^ (1 << bit_loc)
                if delete_bit_mask in bitmask_2_idx:
                    dsu.union(bitmask_2_idx[delete_bit_mask], i)
                for k in range(0, bit_loc):
                    if (1 << k) & delete_bit_mask != 0:
                        continue
                    alter_bit_mask = delete_bit_mask + (1 << k)
                    if alter_bit_mask in bitmask_2_idx:
                        dsu.union(bitmask_2_idx[alter_bit_mask], i)
            bitmask_2_idx[bit_mask] = i
        group_num = 0
        largest_group_size = 0
        for i in range(n):
            group_id = dsu.find(i)
            if group_id == i:
                group_num += 1
                largest_group_size = max(largest_group_size, dsu.size[group_id])
        
        return [group_num, largest_group_size]
        


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