LeetCode 721 - Accounts Merge
| Union Find | DFS | BFS | HashMap | Sorting |
Problem Link on Leetcode: Accounts Merge
Problem Description:
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.
Implementation: Union Find / DisjointSet Approach.
If you need to brush up your knowledge about disjoint set, below is the link:
Union By Rank and Path Compression in Union-Find (Disjoint Sets)
Example 3:
Input: List of accounts with names and emails.
accounts =
[
["JOHN", "J1", "J2"],
["JOHN", "J1", "J3"],
["JOHN", "J3", "J4"],
["JOHN", "J5"],
["MARRY", "M1"]
]
Steps:
- Emails to Account Map:
{
M1=MARRY,
J1=JOHN,
J2=JOHN,
J3=JOHN,
J4=JOHN,
J5=JOHN
}
- Initialization: Emails to Parent Map after initialization of Disjoint Set:
{
J2=J2,
J3=J3,
J4=J4,
J5=J5,
M1=M1,
J1=J1
}
- Emails to parent Map after Union Find:
{
J2=J1,
J3=J1,
J4=J1,
J5=J5,
M1=M1,
J1=J1
}
- Merged Accounts without account name (For each email find it’s parent and add the email to parent’s list of emails):
{
M1=[M1],
J1=[J1, J2, J3, J4],
J5=[J5]
}
- Merged Accounts with account name, Map of Parent/Representative with their emails:
{
M1=[MARRY, M1],
J1=[JOHN, J1, J2, J3, J4],
J5=[JOHN, J5]
}
- Final Merged Accounts List:
[
[MARRY, M1],
[JOHN, J1, J2, J3, J4],
[JOHN, J5]
]
Example 4:
Input: List of accounts with names and emails.
accounts =
[
["ALICE", "A1", "A2"],
["BOB", "B2", "B3"],
["ALICE", "A3", "A4"]
]
Steps:
- Emails to Account Map:
{
A1=ALICE,
B2=BOB,
A2=ALICE,
B3=BOB,
A3=ALICE,
A4=ALICE
}
- Initialization/Emails to Parent Map after initialization of Disjoint Set:
{
A1=A1,
B2=B2,
A2=A2,
B3=B3,
A3=A3,
A4=A4
}
- Emails to parent Map after Union Find:
{
A1=A1,
B2=B2,
A2=A1,
B3=B2,
A3=A3,
A4=A3
}
- Merged Accounts without account name (For each email find it’s parent and add the email to parent’s list of emails):
{
A1=[A1, A2],
B2=[B2, B3],
A3=[A3, A4]
}
- Merged Accounts with account name, Map of Parent/Representative with their emails:
{
A1=[ALICE, A1, A2],
B2=[BOB, B2, B3],
A3=[ALICE, A3, A4]
}
- Final Merged Accounts List:
[
[ALICE, A1, A2],
[BOB, B2, B3],
[ALICE, A3, A4]]
]
Example 5:
Input: List of accounts with names and emails.
accounts =
[
["David","d0","d1"],
["David","d3","d4"],
["David","d4","d5"],
["David","d2","d3"],
["David","d1","d2"]
]
Steps:
- Emails to Account Map:
Map<String, String> emailToNameMap = new HashMap<>();
// Map emails to account names
for (List<String> account : accounts) {
String accountName = account.get(0);
for (int i = 1; i < account.size(); i++) {
String email = account.get(i);
emailToNameMap.putIfAbsent(email, accountName);
}
}
{
d4=David,
d5=David,
d0=David,
d1=David,
d2=David,
d3=David
}
- Initialization: Emails to Parent Map after initialization of Disjoint Set:
public DisjointSet(Set<String> emails) {
int n = emails.size();
parent = new HashMap<>(n);
rank = new HashMap<>(n);
for (String email : emails) {
parent.put(email, email);
rank.put(email, 0);
}
}
{
d4=d4,
d5=d5,
d0=d0,
d1=d1,
d2=d2,
d3=d3
}
- Emails to parent Map after Union Find:
// union accounts
for (List<String> account : accounts) {
String firstEmail = account.get(1);
for (int i = 2; i < account.size(); i++) {
ds.union(firstEmail, account.get(i));
}
}
{
d4=d3,
d5=d3,
d0=d0,
d1=d0,
d2=d3,
d3=d0
}
- Merged Accounts without account name (For each email find it’s parent and add the email to parent’s list of emails):
Map<String, List<String>> mergedEmails = new HashMap<>();
for(String email: emailToNameMap.keySet()){
String parent = ds.find(email);
mergedEmails.putIfAbsent(parent, new ArrayList<String>());
mergedEmails.get(parent).add(email);
}
{
d0=[d4, d5, d0, d1, d2, d3]
}
- Merged Accounts with account name, Map of Parent/Representative with their emails:
mergedEmails.forEach((parent, emails) -> {
Collections.sort(emails);
emails.add(0, emailToNameMap.get(parent));
mergedAccounts.add(emails);
});
{
d0=[David, d0, d1, d2, d3, d4, d5]
}
- Final Merged Accounts List:
[
[David, d0, d1, d2, d3, d4, d5]
]
Edge cases:
Accounts with circular relationships (e.g., [“A”, “B”], [“B”, “C”], [“C”, “A”]).
Account with Single email.
Full Implementation:
class Solution {
public class DisjointSet {
private Map<String, String> parent = new HashMap<>();
private Map<String, Integer> rank;
public DisjointSet(Set<String> emails) {
int n = emails.size();
parent = new HashMap<>(n);
rank = new HashMap<>(n);
for (String email : emails) {
parent.put(email, email);
// Default value of rank is null since we are using HashMap, initialize to zero.
rank.put(email, 0);
}
}
public String find(String x) {
String rep = x;
if (!parent.get(x).equals(x)) {
rep = find(parent.get(x));
parent.put(x, rep); // Path compression
}
return rep;
}
public void union(String x, String y) {
String p1 = find(x);
String p2 = find(y);
if (!p1.equalsIgnoreCase(p2)) {
if (rank.get(p1) > rank.get(p2)) {
parent.put(p2, p1);
} else if (rank.get(p1) < rank.get(p2)) {
parent.put(p1, p2);
} else {
parent.put(p2, p1);
rank.merge(p1, 1, Integer::sum);
}
}
}
}
public List<List<String>> accountsMerge(List<List<String>> accounts) {
List<List<String>> mergedAccounts = new ArrayList<>();
Map<String, String> emailToNameMap = new HashMap<>();
// Map emails to account names
for (List<String> account : accounts) {
String accountName = account.get(0);
for (int i = 1; i < account.size(); i++) {
String email = account.get(i);
emailToNameMap.putIfAbsent(email, accountName);
}
}
DisjointSet ds = new DisjointSet(emailToNameMap.keySet());
// union accounts
for (List<String> account : accounts) {
String firstEmail = account.get(1);
for (int i = 2; i < account.size(); i++) {
ds.union(firstEmail, account.get(i));
}
}
Map<String, List<String>> mergedEmails = new HashMap<>();
//Merged Accounts
for(String email: emailToNameMap.keySet()){
String parent = ds.find(email);
mergedEmails.putIfAbsent(parent, new ArrayList<String>());
mergedEmails.get(parent).add(email);
}
//Sort emails, add account name and accounts to final list
mergedEmails.forEach((parent, emails) -> {
Collections.sort(emails);
emails.add(0, emailToNameMap.get(parent));
mergedAccounts.add(emails);
});
return mergedAccounts;
}
}
Time Complexity:
O(nk log nk + nkα(n)) = O(nk log nk),
n is number of accounts and k = maximum number of emails in accounts[i]
α(n) is inverse Ackermann function which grows very slowly, almost constant time.
Space Complexity:
O(n + nk) = O(nk)
Author: Mohammad J Iqbal
