Given a list of accounts where each element accounts [ i ] is a list of strings, where the first element account [ 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.
Note: Accounts themselves can be returned in any order.
 
Example 1:
Input:
N = 4
accounts [ ] =
[["John","johnsmith@mail.com","john_newyork@mail.com"],
["John","johnsmith@mail.com","john00@mail.com"],
["Mary","mary@mail.com"],
["John","johnnybravo@mail.com"]]
Output:
[["John","john00@mail.com","john_newyork@mail.com", "johnsmith@mail.com"],
["Mary","mary@mail.com"],
["John","johnnybravo@mail.com"]]
Explanation:
The first and second John's are the same person as
they have the common email "johnsmith@mail.com".
The third John and Mary are different people as none
of their email addresses are used by other accounts.
We could return these arrays in any order, for example,
the answer [['Mary', 'mary@mail.com'],
['John', 'johnnybravo@mail.com'],
['John', 'john00@mail.com', 'john_newyork@mail.com',
'johnsmith@mail.com']]
would still be accepted.
 
Example 2:
Input:
N = 5
accounts [ ] =
[["Gabe","Gabe00@m.co","Gabe3@m.co","Gabe1@m.co"],
["Kevin","Kevin3@m.co","Kevin5@m.co","Kevin0@m.co"],
["Ethan","Ethan5@m.co","Ethan4@m.co","Ethan0@m.co"],
["Hanzo","Hanzo3@m.co","Hanzo1@m.co","Hanzo0@m.co"],
["Fern","Fern5@m.co","Fern1@m.co","Fern0@m.co"]]
Output:
[["Ethan","Ethan0@m.co","Ethan4@m.co","Ethan5@m.co"],
["Gabe","Gabe0@m.co","Gabe1@m.co","Gabe3@m.co"],
["Hanzo","Hanzo0@m.co","Hanzo1@m.co","Hanzo3@m.co"],
["Kevin","Kevin0@m.co","Kevin3@m.co","Kevin5@m.co"],
["Fern","Fern0@m.co","Fern1@m.co","Fern5@m.co"]]
Explanation:
We don't have any common emails in any of the users.
We just sorted the emails of each person and we return a array of emails.(The details can be returned in any order).
 
Your Task:
You don't need to read or print anything. Your task is to complete the function accountsMerge() which takes a list of list of strings details representing accounts  as parameter and returns a list of list of strings denoting the details after merging.
Expected Time Complexity: O(N*M*logN) - where N is the size of accounts and M is the size of number of strings for a name.
Expected Auxiliary Space: O(N*M) - where N is the size of accounts and M is the size of number of strings for a name.
 
Constraints:
	1 <= N <= 1000
	2 <= accounts[ i ].size <= 10
	1 <= accounts[ i ][ j ].size <= 30
	accounts[i][0] consists of English letters.

Difficulty: MEDIUM_HARD
Tags: Graph algorithms

Graph traversal

String algorithms

Spanning trees

Data structures
Skill types: Data structures
Expected Time Complexity: O(N*M*logN) - where N is the size of accounts and M is the size of number of strings for a name.
Expected Auxiliary Space: O(N*M) - where N is the size of accounts and M is the size of number of strings for a name.

Starter code:
#User function Template for python3



class Solution:

    def accountsMerge(self, accounts):

        # Code here

Solutions:
import collections

class UF:

	def __init__(self, N):
		self.parents = list(range(N))

	def union(self, child, parent):
		self.parents[self.find(child)] = self.find(parent)

	def find(self, x):
		if x != self.parents[x]:
			self.parents[x] = self.find(self.parents[x])
		return self.parents[x]

class Solution:

	def accountsMerge(self, accounts):
		uf = UF(len(accounts))
		ownership = {}
		for (i, (_, *emails)) in enumerate(accounts):
			for email in emails:
				if email in ownership:
					uf.union(i, ownership[email])
				ownership[email] = i
		ans = collections.defaultdict(list)
		for (email, owner) in ownership.items():
			ans[uf.find(owner)].append(email)
		return [[accounts[i][0]] + sorted(emails) for (i, emails) in ans.items()]


import collections

class Solution:

	def dfs(self, graph, node, visit):
		visit.add(node)
		for nei in graph[node]:
			if nei not in visit:
				self.dfs(graph, nei, visit)
		self.res.append(node)

	def accountsMerge(self, accounts):
		graph = collections.defaultdict(set)
		for account in accounts:
			for email in account[1:]:
				graph[account[1]].add(email)
				graph[email].add(account[1])
		visit = set()
		ans = []
		for account in accounts:
			name = account[0]
			for email in account[1:]:
				if email not in visit:
					self.res = []
					self.dfs(graph, email, visit)
					ans.append([name] + sorted(self.res))
		return ans