[link](https://leetcode.com/problems/accounts-merge/)

---

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

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.

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 in sorted order. The accounts themselves can be returned in any order.

#### Example 1:

Input: accounts = [["John","[e-mail]","[e-mail]"],["John","[e-mail]","[e-mail]"],["Mary","[e-mail]"],["John","[e-mail]"]]
Output: [["John","[e-mail]","[e-mail]","[e-mail]"],["Mary","[e-mail]"],["John","[e-mail]"]]
Explanation:
The first and second John's are the same person as they have the common email "[e-mail]".
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 [['Mary', '[e-mail]'], ['John', '[e-mail]'], 
['John', '[e-mail]', '[e-mail]', '[e-mail]']] would still be accepted.

#### Example 2:

Input: accounts = [["Gabe","[e-mail]","[e-mail]","[e-mail]"],["Kevin","[e-mail]","[e-mail]","[e-mail]"],["Ethan","[e-mail]","[e-mail]","[e-mail]"],["Hanzo","[e-mail]","[e-mail]","[e-mail]"],["Fern","[e-mail]","[e-mail]","[e-mail]"]]
Output: [["Ethan","[e-mail]","[e-mail]","[e-mail]"],["Gabe","[e-mail]","[e-mail]","[e-mail]"],["Hanzo","[e-mail]","[e-mail]","[e-mail]"],["Kevin","[e-mail]","[e-mail]","[e-mail]"],["Fern","[e-mail]","[e-mail]","[e-mail]"]]

#### Constraints:

- 1 <= accounts.length <= 1000
- 2 <= accounts[i].length <= 10
- 1 <= accounts[i][j].length <= 30
- accounts[i][0] consists of English letters.
- accounts[i][j] (for j > 0) is a valid email.

---

UnionFind class: This class implements the Union-Find data structure. It has methods to initialize the data structure, find the parent of a node while performing path compression, and perform the union operation.

find(x): This method finds the parent of node x while performing path compression. It returns the root of the tree that node x belongs to, and it also updates the parent of x to point directly to the root, which helps optimize future find operations.

union(x1, x2): This method performs the union operation between nodes x1 and x2. It first finds the roots of the trees to which x1 and x2 belong using the find method. If the roots are the same, it means the nodes are already in the same group, and the method returns False. Otherwise, it performs the union by updating the parent and rank arrays and returns True.

Main algorithm: The accountsMerge method takes a list of accounts as input. It first initializes a Union-Find data structure with the number of accounts. It then creates a dictionary emailToAcc to map each email to its corresponding account index. For each account, it iterates through the emails and checks if the email is already in emailToAcc. If it is, it performs the union between the current account index and the account index associated with the email. If it is not, it adds the email and its corresponding account index to emailToAcc.

After processing all accounts, the algorithm groups the accounts by their root parent using a defaultdict emailGroup. It collects all emails with the same root parent into the same group.

Finally, the algorithm constructs the result by iterating through the email groups in emailGroup. For each group, it collects the account names and sorts the emails before adding them to the result list.

#### Solution 1

class UnionFind():
    def __init__(self, n):
        self.par = [i for i in range(n)]
        self.rank = [0] * n
    

    def find(self, x):
        while x != self.par[x]:
            self.par[x] = self.par[self.par[x]]
            x = self.par[x]
        return x


    def union(self, x1, x2):
        p1 = self.find(x1)
        p2 = self.find(x2)
        if p1 == p2:
            return False
        if self.rank[p1] > self.rank[p2]:
            self.par[p2] = p1
            self.rank[p1] += self.rank[p2]
        else:
            self.par[p1] = p2
            self.rank[p2] += self.rank[p1]
        return True


class Solution:
    def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
        uf = UnionFind(len(accounts))
        emailToAcc = {}
        for i, a in enumerate(accounts):
            for e in a[1:]:
                if e in emailToAcc:
                    uf.union(i, emailToAcc[e])
                else:
                    emailToAcc[e] = i
        
        emailGroup = defaultdict(list)
        for e, i in emailToAcc.items():
            leader = uf.find(i)
            emailGroup[leader].append(e)

        res = []
        for i, emails in emailGroup.items():
            name = accounts[i][0]
            res.append([name] + sorted(emailGroup[i]))

        return res

        

The time complexity of the algorithm is O(nmlog(m)), where n is the number of accounts and m is the average number of emails per account. The Union-Find operations are close to O(1) on average, but the overall complexity is influenced by the sorting step.

The space complexity is O(n*m), where n is the number of accounts and m is the average number of emails per account, as we use the Union-Find data structure and the emailToAcc and emailGroup dictionaries to store the data.