Accounts Merge

Accounts Merge

What is this problem about?

The Accounts Merge coding problem gives you a list of accounts where each account has a name and a list of emails. The catch? The same person can have multiple accounts, but only if there is at least one common email between them. You need to merge these accounts so that each person has only one consolidated account with all their unique emails sorted alphabetically.

Why is this asked in interviews?

This is a favorite at Microsoft, Meta, and Pinterest because it is a classic Graph connectivity problem. It tests your ability to model real-world data as a graph where emails are nodes and an account connects those nodes. It evaluates your proficiency with Union-Find or DFS/BFS.

Algorithmic pattern used

The most common approach is the Union-Find interview pattern (Disjoint Set Union). Each email is mapped to an ID. If two emails appear in the same account, you "Union" them. After processing all accounts, all emails belonging to the same person will point to the same root representative. Alternatively, you can use Depth-First Search (DFS) to find connected components in the email graph.

Example explanation

• Account 1: John, ["a@m.com", "b@m.com"]
• Account 2: John, ["c@m.com", "b@m.com"]
• Account 3: Mary, ["d@m.com"]

Because Account 1 and 2 both share b@m.com, they belong to the same John. Merged Result:

• John: ["a@m.com", "b@m.com", "c@m.com"] (sorted)
• Mary: ["d@m.com"]

Common mistakes candidates make

• Merging by Name: Assuming everyone with the name "John" is the same person. The problem states that the name is just a label; the email is the only unique identifier.
• Inefficient Sorting: Sorting the emails too early. You should only sort once per merged account at the very end.
• String Manipulation Slowness: Using strings directly in Union-Find. It's much faster to map each email to an integer ID first.

Interview preparation tip

When a problem asks you to "group" or "merge" items based on common relationships, your mind should immediately go to Union-Find. It’s the most efficient way to handle dynamic connectivity in a graph.