Accounts Merge

Problem

Given a list of accounts where each element accounts[i] is a list of strings, where the first element is a name and the rest are emails, merge accounts that share the same email. Return the merged accounts in any order.

Examples

The brute-force approach

Use DFS/BFS: build a graph where emails are nodes and two emails in the same account are connected. Find connected components; each component is one merged account.

# Build adjacency list from emails in same account
for account in accounts:
    for i in range(1, len(account)):
        for j in range(i+1, len(account)):
            adj[account[i]].append(account[j])
            adj[account[j]].append(account[i])
# DFS to find connected components
visited = set()
result = []
for email in all_emails:
    if email not in visited:
        component = dfs(email, visited, adj)
        name = email_to_name[email]
        result.append([name] + sorted(component))

Building all pairs within each account is O(emails_per_account²). Union-Find connects to the first email of each account in one pass.

Spotting the pattern

This is a Union-Find problem. The key question to ask yourself:

Two accounts should be merged if they share any email. If you pick one "anchor" email per account and union all others to it, when will two anchors end up in the same component?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.