leetcode721

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721. Accounts Merge

Given a list accounts, 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 email that is common 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:

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Input: 
accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
Explanation: 
The first and third John's are the same person as they have the common email "johnsmith@mail.com".
The second 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', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.

Note:

The length of accounts will be in the range [1, 1000].
The length of accounts[i] will be in the range [1, 10].
The length of accounts[i][j] will be in the range [1, 30].

Idea

Union-find the emails. The emails in an account belong to one group. Since, we can not directly perform uf operations on the strs, so we need mapping them between int array.

Code

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class UnionFind(object):
    def __init__(self, n):
        self.parent = [i for i in range(n)]
        self.rank = [0] * n
        
    def union(self, n, m):
        p, q = self.find(n), self.find(m)
        if p != q:
            if self.rank[p] > self.rank[q]:
                self.parent[q] = p
            else:
                self.parent[p] = q
                if self.rank[p] == self.rank[q]:
                    self.rank[q] += 1
        
    def find(self, n):
        if self.parent[n] != n:
            return self.find(self.parent[n])
        return n

class Solution(object):
    def accountsMerge(self, accounts):
        """
        :type accounts: List[List[str]]
        :rtype: List[List[str]]
        """
        # init
        eid = 0     # email id
        e2n = {}    # email to name
        e2i = {}    # email to email id
        i2e = {}    # email id to email
        for account in accounts:
            name = account[0]
            for email in account[1:]:
                if not (email in e2i):
                    e2n[email] = name
                    e2i[email] = eid
                    i2e[eid] = email
                    eid += 1

        # union find
        uf = UnionFind(eid + 1)
        for account in accounts:
            first_eid = e2i[account[1]]
            for email in account[2:]:
                uf.union(first_eid, e2i[email])
        
        # prev steps to construct answer:
        # find the all the parent emails as well as their children in a group
        p2c = collections.defaultdict(set)    # parent email to children emails
        for account in accounts:
            for email in account[1:]:
                parent_email_id = uf.find(e2i[email])
                parent_email = i2e[parent_email_id]
                p2c[parent_email].add(email)
        
        # construct the answers
        res = []
        for parent in p2c:
            tmp = []
            name = e2n[parent]
            tmp = [name]
            tmp += sorted(list(p2c[parent]))
            res.append(tmp)
            
        return res