Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
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 <= 10002 <= accounts[i].length <= 101 <= accounts[i][j].length <= 30accounts[i][0] consists of English letters.accounts[i][j] (for j > 0) is a valid email.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Hash Map · Union-Find
[["John","johnsmith@mail.com","john_newyork@mail.com"],["John","johnsmith@mail.com","john00@mail.com"],["Mary","mary@mail.com"],["John","johnnybravo@mail.com"]]
[["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"]]
redundant-connection)sentence-similarity)sentence-similarity-ii)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #721: Accounts Merge
class UnionFind {
private final int[] p;
private final int[] size;
public UnionFind(int n) {
p = new int[n];
size = new int[n];
for (int i = 0; i < n; ++i) {
p[i] = i;
size[i] = 1;
}
}
public int find(int x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
public boolean union(int a, int b) {
int pa = find(a), pb = find(b);
if (pa == pb) {
return false;
}
if (size[pa] > size[pb]) {
p[pb] = pa;
size[pa] += size[pb];
} else {
p[pa] = pb;
size[pb] += size[pa];
}
return true;
}
}
class Solution {
public List<List<String>> accountsMerge(List<List<String>> accounts) {
int n = accounts.size();
UnionFind uf = new UnionFind(n);
Map<String, Integer> d = new HashMap<>();
for (int i = 0; i < n; ++i) {
for (int j = 1; j < accounts.get(i).size(); ++j) {
String email = accounts.get(i).get(j);
if (d.containsKey(email)) {
uf.union(i, d.get(email));
} else {
d.put(email, i);
}
}
}
Map<Integer, Set<String>> g = new HashMap<>();
for (int i = 0; i < n; ++i) {
int root = uf.find(i);
g.computeIfAbsent(root, k -> new HashSet<>())
.addAll(accounts.get(i).subList(1, accounts.get(i).size()));
}
List<List<String>> ans = new ArrayList<>();
for (var e : g.entrySet()) {
List<String> emails = new ArrayList<>(e.getValue());
Collections.sort(emails);
ans.add(new ArrayList<>());
ans.get(ans.size() - 1).add(accounts.get(e.getKey()).get(0));
ans.get(ans.size() - 1).addAll(emails);
}
return ans;
}
}
// Accepted solution for LeetCode #721: Accounts Merge
type unionFind struct {
p, size []int
}
func newUnionFind(n int) *unionFind {
p := make([]int, n)
size := make([]int, n)
for i := range p {
p[i] = i
size[i] = 1
}
return &unionFind{p, size}
}
func (uf *unionFind) find(x int) int {
if uf.p[x] != x {
uf.p[x] = uf.find(uf.p[x])
}
return uf.p[x]
}
func (uf *unionFind) union(a, b int) bool {
pa, pb := uf.find(a), uf.find(b)
if pa == pb {
return false
}
if uf.size[pa] > uf.size[pb] {
uf.p[pb] = pa
uf.size[pa] += uf.size[pb]
} else {
uf.p[pa] = pb
uf.size[pb] += uf.size[pa]
}
return true
}
func accountsMerge(accounts [][]string) (ans [][]string) {
n := len(accounts)
uf := newUnionFind(n)
d := make(map[string]int)
for i := 0; i < n; i++ {
for _, email := range accounts[i][1:] {
if j, ok := d[email]; ok {
uf.union(i, j)
} else {
d[email] = i
}
}
}
g := make(map[int]map[string]struct{})
for i := 0; i < n; i++ {
root := uf.find(i)
if _, ok := g[root]; !ok {
g[root] = make(map[string]struct{})
}
for _, email := range accounts[i][1:] {
g[root][email] = struct{}{}
}
}
for root, s := range g {
emails := []string{}
for email := range s {
emails = append(emails, email)
}
sort.Strings(emails)
account := append([]string{accounts[root][0]}, emails...)
ans = append(ans, account)
}
return
}
# Accepted solution for LeetCode #721: Accounts Merge
class UnionFind:
def __init__(self, n):
self.p = list(range(n))
self.size = [1] * n
def find(self, x):
if self.p[x] != x:
self.p[x] = self.find(self.p[x])
return self.p[x]
def union(self, a, b):
pa, pb = self.find(a), self.find(b)
if pa == pb:
return False
if self.size[pa] > self.size[pb]:
self.p[pb] = pa
self.size[pa] += self.size[pb]
else:
self.p[pa] = pb
self.size[pb] += self.size[pa]
return True
class Solution:
def accountsMerge(self, accounts: List[List[str]]) -> List[List[str]]:
uf = UnionFind(len(accounts))
d = {}
for i, (_, *emails) in enumerate(accounts):
for email in emails:
if email in d:
uf.union(i, d[email])
else:
d[email] = i
g = defaultdict(set)
for i, (_, *emails) in enumerate(accounts):
root = uf.find(i)
g[root].update(emails)
return [[accounts[root][0]] + sorted(emails) for root, emails in g.items()]
// Accepted solution for LeetCode #721: Accounts Merge
struct Solution;
use rustgym_util::*;
use std::collections::BTreeMap;
use std::collections::BTreeSet;
use std::collections::HashMap;
impl Solution {
fn accounts_merge(accounts: Vec<Vec<String>>) -> Vec<Vec<String>> {
let n = accounts.len();
let mut btm: BTreeMap<&str, &str> = BTreeMap::new();
for i in 0..n {
let m = accounts[i].len();
let name = &accounts[i][0];
for j in 1..m {
let email = &accounts[i][j];
btm.insert(email, name);
}
}
let count = btm.len();
let mut uf = UnionFind::new(count);
let mut owners = vec![];
let mut emails = vec![];
let mut ids = HashMap::new();
for (i, (email, name)) in btm.into_iter().enumerate() {
emails.push(email.to_string());
owners.push(name.to_string());
ids.insert(email.to_string(), i);
}
for i in 0..n {
let m = accounts[i].len();
for j in 2..m {
let email_a = &accounts[i][j - 1];
let email_b = &accounts[i][j];
let id_a = ids[email_a];
let id_b = ids[email_b];
uf.union(id_a, id_b);
}
}
let mut res: Vec<Vec<String>> = vec![];
let mut hm: HashMap<usize, BTreeSet<usize>> = HashMap::new();
for i in 0..count {
let group_id = uf.find(i);
hm.entry(group_id).or_default().insert(i);
}
for (group_id, ids) in hm.into_iter() {
let mut v: Vec<String> = vec![owners[group_id].clone()];
for id in ids {
v.push(emails[id].to_string());
}
res.push(v);
}
res
}
}
#[test]
fn test() {
let accounts = vec![
vec_string!["John", "johnsmith@mail.com", "john00@mail.com"],
vec_string!["John", "johnnybravo@mail.com"],
vec_string!["John", "johnsmith@mail.com", "john_newyork@mail.com"],
vec_string!["Mary", "mary@mail.com"],
];
let mut ans = vec![
vec_string![
"John",
"john00@mail.com",
"john_newyork@mail.com",
"johnsmith@mail.com"
],
vec_string!["John", "johnnybravo@mail.com"],
vec_string!["Mary", "mary@mail.com"],
];
let mut res = Solution::accounts_merge(accounts);
res.sort_unstable();
ans.sort_unstable();
assert_eq!(ans, res);
}
// Accepted solution for LeetCode #721: Accounts Merge
class UnionFind {
private p: number[];
private size: number[];
constructor(n: number) {
this.p = new Array(n);
this.size = new Array(n);
for (let i = 0; i < n; ++i) {
this.p[i] = i;
this.size[i] = 1;
}
}
find(x: number): number {
if (this.p[x] !== x) {
this.p[x] = this.find(this.p[x]);
}
return this.p[x];
}
union(a: number, b: number): boolean {
let pa = this.find(a),
pb = this.find(b);
if (pa === pb) {
return false;
}
if (this.size[pa] > this.size[pb]) {
this.p[pb] = pa;
this.size[pa] += this.size[pb];
} else {
this.p[pa] = pb;
this.size[pb] += this.size[pa];
}
return true;
}
}
function accountsMerge(accounts: string[][]): string[][] {
const n = accounts.length;
const uf = new UnionFind(n);
const d = new Map<string, number>();
for (let i = 0; i < n; ++i) {
for (let j = 1; j < accounts[i].length; ++j) {
const email = accounts[i][j];
if (d.has(email)) {
uf.union(i, d.get(email)!);
} else {
d.set(email, i);
}
}
}
const g = new Map<number, Set<string>>();
for (let i = 0; i < n; ++i) {
const root = uf.find(i);
if (!g.has(root)) {
g.set(root, new Set<string>());
}
const emailSet = g.get(root)!;
for (let j = 1; j < accounts[i].length; ++j) {
emailSet.add(accounts[i][j]);
}
}
const ans: string[][] = [];
for (const [root, emails] of g.entries()) {
const emailList = Array.from(emails).sort();
const mergedAccount = [accounts[root][0], ...emailList];
ans.push(mergedAccount);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.
With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: Zero-count keys stay in map and break distinct/count constraints.
Usually fails on: Window/map size checks are consistently off by one.
Fix: Delete keys when count reaches zero.