Approach #1: DepthFirst Search [Accepted]
Intuition
Draw an edge between two emails if they occur in the same account. The problem comes down to finding connected components of this graph.
Algorithm
For each account, draw the edge from the first email to all other emails. Additionally, we'll remember a map from emails to names on the side. After finding each connected component using a depthfirst search, we'll add that to our answer.
Python
class Solution(object): def accountsMerge(self, accounts): em_to_name = {} graph = collections.defaultdict(set) for acc in accounts: name = acc[0] for email in acc[1:]: graph[acc[1]].add(email) graph[email].add(acc[1]) em_to_name[email] = name seen = set() ans = [] for email in graph: if email not in seen: seen.add(email) stack = [email] component = [] while stack: node = stack.pop() component.append(node) for nei in graph[node]: if nei not in seen: seen.add(nei) stack.append(nei) ans.append([em_to_name[email]] + sorted(component)) return ans
Java
class Solution { public List<List<String>> accountsMerge(List<List<String>> accounts) { Map<String, String> emailToName = new HashMap(); Map<String, ArrayList<String>> graph = new HashMap(); for (List<String> account: accounts) { String name = ""; for (String email: account) { if (name == "") { name = email; continue; } graph.computeIfAbsent(email, x> new ArrayList<String>()).add(account.get(1)); graph.computeIfAbsent(account.get(1), x> new ArrayList<String>()).add(email); emailToName.put(email, name); } } Set<String> seen = new HashSet(); List<List<String>> ans = new ArrayList(); for (String email: graph.keySet()) { if (!seen.contains(email)) { seen.add(email); Stack<String> stack = new Stack(); stack.push(email); List<String> component = new ArrayList(); while (!stack.empty()) { String node = stack.pop(); component.add(node); for (String nei: graph.get(node)) { if (!seen.contains(nei)) { seen.add(nei); stack.push(nei); } } } Collections.sort(component); component.add(0, emailToName.get(email)); ans.add(component); } } return ans; } }
Complexity Analysis

Time Complexity: , where is the length of
accounts[i]
. Without the log factor, this is the complexity to build the graph and search for each component. The log factor is for sorting each component at the end. 
Space Complexity: , the space used by our graph and our search.
Approach #2: UnionFind [Accepted]
Intuition
As in Approach #1, our problem comes down to finding the connected components of a graph. This is a natural fit for a Disjoint Set Union (DSU) structure.
Algorithm
As in Approach #1, draw edges between emails if they occur in the same account. For easier interoperability between our DSU template, we will map each email to some integer index by using emailToID
. Then, dsu.find(email)
will tell us a unique id representing what component that email is in.
For more information on DSU, please look at Approach #2 in the article here. For brevity, the solutions showcased below do not use unionbyrank.
Python
class DSU: def __init__(self): self.p = range(10001) def find(self, x): if self.p[x] != x: self.p[x] = self.find(self.p[x]) return self.p[x] def union(self, x, y): self.p[self.find(x)] = self.find(y) class Solution(object): def accountsMerge(self, accounts): dsu = DSU() em_to_name = {} em_to_id = {} i = 0 for acc in accounts: name = acc[0] for email in acc[1:]: em_to_name[email] = name if email not in em_to_id: em_to_id[email] = i i += 1 dsu.union(em_to_id[acc[1]], em_to_id[email]) ans = collections.defaultdict(list) for email in em_to_name: ans[dsu.find(em_to_id[email])].append(email) return [[em_to_name[v[0]]] + sorted(v) for v in ans.values()]
Java
class Solution { public List<List<String>> accountsMerge(List<List<String>> accounts) { DSU dsu = new DSU(); Map<String, String> emailToName = new HashMap(); Map<String, Integer> emailToID = new HashMap(); int id = 0; for (List<String> account: accounts) { String name = ""; for (String email: account) { if (name == "") { name = email; continue; } emailToName.put(email, name); if (!emailToID.containsKey(email)) { emailToID.put(email, id++); } dsu.union(emailToID.get(account.get(1)), emailToID.get(email)); } } Map<Integer, List<String>> ans = new HashMap(); for (String email: emailToName.keySet()) { int index = dsu.find(emailToID.get(email)); ans.computeIfAbsent(index, x> new ArrayList()).add(email); } for (List<String> component: ans.values()) { Collections.sort(component); component.add(0, emailToName.get(component.get(0))); } return new ArrayList(ans.values()); } } class DSU { int[] parent; public DSU() { parent = new int[10001]; for (int i = 0; i <= 10000; ++i) parent[i] = i; } public int find(int x) { if (parent[x] != x) parent[x] = find(parent[x]); return parent[x]; } public void union(int x, int y) { parent[find(x)] = find(y); } }
Complexity Analysis

Time Complexity: , where , and is the length of
accounts[i]
. If we used unionbyrank, this complexity improves to , where is the InverseAckermann function. 
Space Complexity: , the space used by our DSU structure.
Analysis written by: @awice.