Accounts Merge Python Solution with Tests

LeetCode 721, Medium. Topics: Array, Hash Table, String, Depth-First Search, Breadth-First Search, Union Find, Sorting. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 13 parametrized pytest cases, and a playground notebook:

lcpy gen -n 721   # by problem number
lcpy gen -s accounts_merge   # by problem name

Problem

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.

Examples

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 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.

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

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(N * M) where N is accounts, M is max emails per account
    # Space: O(N * M)
    def accounts_merge(self, accounts: list[list[str]]) -> list[list[str]]:
        email_to_accounts: dict[str, list[int]] = {}

        for i, account in enumerate(accounts):
            for email in account[1:]:
                if email not in email_to_accounts:
                    email_to_accounts[email] = []
                email_to_accounts[email].append(i)

        visited: set[int] = set()
        result = []

        def dfs(account_idx: int, emails: set[str]) -> None:
            if account_idx in visited:
                return
            visited.add(account_idx)

            for email in accounts[account_idx][1:]:
                emails.add(email)
                for neighbor_idx in email_to_accounts[email]:
                    dfs(neighbor_idx, emails)

        for i in range(len(accounts)):
            if i in visited:
                continue

            emails: set[str] = set()
            dfs(i, emails)
            result.append([accounts[i][0], *sorted(emails)])

        return result

Complexity

Time	Space
O(N * M) where N is accounts, M is max emails per account	O(N * M)

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Accounts Merge Python Solution with Tests

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