LeetCode 721. Accounts Merge

Description

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","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 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.

Example 2:

Input: accounts = [["Gabe","Gabe0@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"]]

Constraints:

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

Explanation

This email merge problem is actually can be treated as a graph problem. We can first build the adjacency relationship between accounts by finding the common emails. Then start from each account, base on the accounts adjacency relationship, we can use a depth-first search to find all the emails linked with that account.

Python Solution

class Solution:
    def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
        results = []
        
        
        email_visited = {}
        
        adjacency_list = []
        for i in range(len(accounts)):
            adjacency_list.append([])
        
                
        for i, account in enumerate(accounts):
            name = account[0]
            
            emails = sorted(account[1:])
            
            for email in emails:
                if email in email_visited:   
                    adjacency_list[email_visited[email]].append(i)
                    adjacency_list[i].append(email_visited[email])
                else:
                    email_visited[email] = i
      
    
        id_visited = set()
        for i, account in enumerate(accounts):       
            if i in id_visited:
                continue
            
            name = account[0]            
            merge_emails = set(account[1:])
            
            self.helper(accounts, adjacency_list, i, id_visited, name, merge_emails)
        
            merge_emails = sorted(list(merge_emails))
            results.append([name] + merge_emails)

        
        return results
     
    def helper(self, accounts, adjacency_list, idx, visited, name, merge_emails):
        if idx in visited:
            return
        
        visited.add(idx)
        
        for adj_idx in adjacency_list[idx]:            
            merge_emails.update(accounts[adj_idx][1:])
                        
            self.helper(accounts, adjacency_list, adj_idx, visited, name, merge_emails)
  • Time Complexity: O(N^2).
  • Space Complexity: O(N).

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