<p>Given a list of <code>accounts</code> where each element <code>accounts[i]</code> is a list of strings, where the first element <code>accounts[i][0]</code> is a name, and the rest of the elements are <strong>emails</strong> representing emails of the account.</p>

<p>Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some common email to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.</p>

<p>After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails <strong>in sorted order</strong>. The accounts themselves can be returned in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> accounts = [[&quot;John&quot;,&quot;johnsmith@mail.com&quot;,&quot;john_newyork@mail.com&quot;],[&quot;John&quot;,&quot;johnsmith@mail.com&quot;,&quot;john00@mail.com&quot;],[&quot;Mary&quot;,&quot;mary@mail.com&quot;],[&quot;John&quot;,&quot;johnnybravo@mail.com&quot;]]
<strong>Output:</strong> [[&quot;John&quot;,&quot;john00@mail.com&quot;,&quot;john_newyork@mail.com&quot;,&quot;johnsmith@mail.com&quot;],[&quot;Mary&quot;,&quot;mary@mail.com&quot;],[&quot;John&quot;,&quot;johnnybravo@mail.com&quot;]]
<strong>Explanation:</strong>
The first and second John&#39;s are the same person as they have the common email &quot;johnsmith@mail.com&quot;.
The third John and Mary are different people as none of their email addresses are used by other accounts.
We could return these lists in any order, for example the answer [[&#39;Mary&#39;, &#39;mary@mail.com&#39;], [&#39;John&#39;, &#39;johnnybravo@mail.com&#39;], 
[&#39;John&#39;, &#39;john00@mail.com&#39;, &#39;john_newyork@mail.com&#39;, &#39;johnsmith@mail.com&#39;]] would still be accepted.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> accounts = [[&quot;Gabe&quot;,&quot;Gabe0@m.co&quot;,&quot;Gabe3@m.co&quot;,&quot;Gabe1@m.co&quot;],[&quot;Kevin&quot;,&quot;Kevin3@m.co&quot;,&quot;Kevin5@m.co&quot;,&quot;Kevin0@m.co&quot;],[&quot;Ethan&quot;,&quot;Ethan5@m.co&quot;,&quot;Ethan4@m.co&quot;,&quot;Ethan0@m.co&quot;],[&quot;Hanzo&quot;,&quot;Hanzo3@m.co&quot;,&quot;Hanzo1@m.co&quot;,&quot;Hanzo0@m.co&quot;],[&quot;Fern&quot;,&quot;Fern5@m.co&quot;,&quot;Fern1@m.co&quot;,&quot;Fern0@m.co&quot;]]
<strong>Output:</strong> [[&quot;Ethan&quot;,&quot;Ethan0@m.co&quot;,&quot;Ethan4@m.co&quot;,&quot;Ethan5@m.co&quot;],[&quot;Gabe&quot;,&quot;Gabe0@m.co&quot;,&quot;Gabe1@m.co&quot;,&quot;Gabe3@m.co&quot;],[&quot;Hanzo&quot;,&quot;Hanzo0@m.co&quot;,&quot;Hanzo1@m.co&quot;,&quot;Hanzo3@m.co&quot;],[&quot;Kevin&quot;,&quot;Kevin0@m.co&quot;,&quot;Kevin3@m.co&quot;,&quot;Kevin5@m.co&quot;],[&quot;Fern&quot;,&quot;Fern0@m.co&quot;,&quot;Fern1@m.co&quot;,&quot;Fern5@m.co&quot;]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= accounts.length &lt;= 1000</code></li>
	<li><code>2 &lt;= accounts[i].length &lt;= 10</code></li>
	<li><code>1 &lt;= accounts[i][j].length &lt;= 30</code></li>
	<li><code>accounts[i][0]</code> consists of English letters.</li>
	<li><code>accounts[i][j] (for j &gt; 0)</code> is a valid email.</li>
</ul>

class UnionFind:
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        # rank is the size of each set
        self.rank = [1] * n
    
    def find(self, x):
        # Find the root parent
        while x != self.parent[x]:
            # Keep searching while res is not equal to its parent
            # Set the current parent to the grandparent (path compression)
            self.parent[x] = self.parent[self.parent[x]]
            x = self.parent[x]
        return x
    
    def union(self, x, y):
        px, py = self.find(x), self.find(y)
        # If they have the same parent
        if px == py: return False
        # Set with fewer nodes will be the child of the set with more nodes
        # Parents rank will increase
        if self.rank[py] > self.rank[px]: 
            self.parent[px] = py
            self.rank[py] += self.rank[px]
        else:
            self.parent[py] = px
            self.rank[px] += self.rank[py]
        return True

class Solution:
    def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
        uf = UnionFind(len(accounts))
        email_to_acc = {} # email -> index of acc

        # Create disjoint sets
        for i, a in enumerate(accounts):
            # Go through each email for each account
            for e in a[1:]:
                if e in email_to_acc:
                    # If the email already exists, then union the current one with the one existing
                    uf.union(i, email_to_acc[e])
                else:
                    # Map email to current account
                    email_to_acc[e] = i
        
        email_group = defaultdict(list) # index of acc -> list of emails
        for e, i in email_to_acc.items():
            leader = uf.find(i) # Get the parent of the current account
            email_group[leader].append(e) # Append the current email to the leader of the current account
        
        # Format the list
        res = []
        for i, emails in email_group.items():
            name_of_acc = accounts[i][0]
            res.append([name_of_acc] + sorted(email_group[i]))
        
        return res


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