<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.ranks = [0 for i in range (n)]
        self.parents = [i for i in range (n)]
        
    def find(self, u):
        if self.parents[u] != u:
            self.parents[u] = self.find(self.parents[u])
        return self.parents[u]
    def union(self, u, v):
        par_u = self.find(u)
        par_v = self.find(v)
        if par_u == par_v:
            return 
        
        if self.ranks[par_u] <= self.ranks[par_v]:
            self.parents[par_u] = par_v
            if self.ranks[par_u] == self.ranks[par_v]:
                self.ranks[par_v] += 1
        else:
            self.parents[par_v] = par_u
    def find_group(self):
        groups = defaultdict(int)
        for i in range (len(self.parents)):
            groups[self.find(i)] += 1
            
        return len(groups), max(groups.values())
class Solution:
    def convert_word(self, word: str):
        ans = 0
        for ch in word:
            val = 1 << (ord(ch) - ord("a"))
            ans += val
            
        return ans
    def find_neighbor(self, num: int):
        neighbors = set()
        for i in range (26):
            neighbors.add(num ^ (1 << i))
            if num & (1 << i):
                for j in range (26):
                    if num & (1 << j) == 0:
                        neighbors.add(num ^ (1 << i) ^ (1 << j))
                        
        return neighbors
    def groupStrings(self, words: List[str]) -> List[int]:
        n = len(words)
        union = UnionFind(n)
        d = defaultdict(int)
        for i in range (n):
            word = words[i]
            num = self.convert_word(word)
            neighbors = self.find_neighbor(num)
            if num in d:
                union.union(d[num], i)
            for neighbor in neighbors:
                if neighbor in d:
                    union.union(d[neighbor], i)
            d[num] = i
            
        ans1, ans2 = union.find_group()
        return [ans1, ans2]
            
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