Introduction

In this blog post, we will explore how to solve the LeetCode problem "721" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's a solution to the account merging problem, along with explanations:

• Build a graph: Represent accounts as nodes and create edges between accounts that share common emails. This allows us to identify connected components, which represent accounts belonging to the same person.
• Union-Find (Disjoint Set Union): Efficiently determine connected components in the graph. Each component represents a merged account.

• Merge and Sort: Iterate through the connected components identified by Union-Find, collect all emails within each component, and sort them. Combine the sorted emails with the account name.

• Time Complexity: O(N M log(M)), where N is the number of accounts and M is the maximum number of emails in an account. The log(M) factor comes from sorting emails. Space Complexity: O(N * M)

Code

    class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])  # Path compression
        return self.parent[x]

    def union(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x != root_y:
            self.parent[root_x] = root_y


def accounts_merge(accounts):
    n = len(accounts)
    uf = UnionFind(n)
    email_to_account = {}

    # Build the Union-Find structure
    for i, account in enumerate(accounts):
        for j in range(1, len(account)):
            email = account[j]
            if email in email_to_account:
                uf.union(i, email_to_account[email])
            else:
                email_to_account[email] = i

    # Merge accounts
    merged_accounts = {}
    for i, account in enumerate(accounts):
        root = uf.find(i)
        if root not in merged_accounts:
            merged_accounts[root] = [account[0]]  # Store the name
        for j in range(1, len(account)):
            email = account[j]
            merged_accounts[root].append(email)

    # Format and sort the results
    result = []
    for root, emails in merged_accounts.items():
        name = emails[0]
        sorted_emails = sorted(list(set(emails[1:])))  # Remove duplicates and sort
        result.append([name] + sorted_emails)

    return result