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AccountsMerge721.java
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AccountsMerge721.java
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/**
* Given a list accounts, 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 email that is common 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", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
* Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'], ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
*
* Explanation:
* The first and third John's are the same person as they have the common email "johnsmith@mail.com".
* The second 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.
*
* Note:
*
* The length of accounts will be in the range [1, 1000].
* The length of accounts[i] will be in the range [1, 10].
* The length of accounts[i][j] will be in the range [1, 30].
*
*/
public class AccountsMerge721 {
public List<List<String>> accountsMerge(List<List<String>> accounts) {
List<List<String>> res = new ArrayList<>();
Map<String, Set<Set<String>>> map = new HashMap<>();
for (List<String> acc: accounts) {
String name = acc.get(0);
if (!map.containsKey(name)) {
Set<String> set = new HashSet<>(acc);
set.remove(name);
Set<Set<String>> setOfSet = new HashSet<>();
setOfSet.add(set);
map.put(name, setOfSet);
} else {
Set<Set<String>> toBeMerged = new HashSet<>();
Set<Set<String>> setOfSet = map.get(name);
Set<String> curr = new HashSet<>(acc);
curr.remove(name);
for (String email: curr) {
Set<String> found = null;
for (Set<String> e: setOfSet) {
if (e.contains(email)) {
found = e;
break;
}
}
if (found != null) {
setOfSet.remove(found);
toBeMerged.add(found);
}
}
for (Set<String> s: toBeMerged) {
curr.addAll(s);
}
setOfSet.add(curr);
}
}
for (Map.Entry<String, Set<Set<String>>> en: map.entrySet()) {
for (Set<String> es: en.getValue()) {
List<String> list = new ArrayList<>(es);
java.util.Collections.sort(list);
list.add(0, en.getKey());
res.add(list);
}
}
return res;
}
/**
* https://leetcode.com/problems/accounts-merge/solution/
*/
int MAX_VAL = 10000;
public List<List<String>> accountsMerge(List<List<String>> accounts) {
Map<String, String> emailToName = new HashMap<>();
Map<String, Integer> emailToId = new HashMap<>();
DSU dsu = initDSU(accounts, emailToName, emailToId);
Map<Integer, List<String>> res = new HashMap();
for (String email: emailToName.keySet()) {
int id = dsu.find(emailToId.get(email));
res.computeIfAbsent(id, x -> new ArrayList()).add(email);
}
for (List<String> emails: res.values()) {
Collections.sort(emails);
emails.add(0, emailToName.get(emails.get(0)));
}
return new ArrayList(res.values());
}
private DSU initDSU(List<List<String>> accounts, Map<String, String> emailToName, Map<String, Integer> emailToId) {
DSU dsu = new DSU(MAX_VAL+1);
int id = 0;
for (List<String> acc: accounts) {
String name = acc.get(0);
for (int i=1; i<acc.size(); i++) {
String email = acc.get(i);
emailToName.putIfAbsent(email, name);
if (!emailToId.containsKey(email)) {
emailToId.put(email, id);
id++;
}
dsu.union(emailToId.get(acc.get(1)), emailToId.get(email));
}
}
return dsu;
}
class DSU {
int[] parent;
public DSU(int size) {
parent = new int[size];
for (int i = 0; i < size; 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);
}
}
/**
* https://leetcode.com/problems/accounts-merge/discuss/109158/Java-Solution-(Build-graph-+-DFS-search)
*/
public List<List<String>> accountsMerge3(List<List<String>> accounts) {
Map<String, String> emailToName = new HashMap<>();
Map<String, Set<String>> graph = initGraph(accounts, emailToName);
Set<String> visited = new HashSet<>();
List<List<String>> res = new LinkedList<>();
for (String email: emailToName.keySet()) {
List<String> list = new LinkedList<>();
if (visited.add(email)) {
dfs(graph, email, visited, list);
Collections.sort(list);
list.add(0, emailToName.get(email));
res.add(list);
}
}
return res;
}
private Map<String, Set<String>> initGraph(List<List<String>> accounts, Map<String, String> emailToName) {
Map<String, Set<String>> graph = new HashMap<>();
for (List<String> acc : accounts) {
String name = acc.get(0);
for (int i = 1; i < acc.size(); i++) {
String email = acc.get(i);
if (!graph.containsKey(acc.get(i))) {
graph.put(email, new HashSet<>());
}
emailToName.put(email, name);
if (i == 1) continue;
graph.get(email).add(acc.get(i-1));
graph.get(acc.get(i-1)).add(email);
}
}
return graph;
}
private void dfs(Map<String, Set<String>> graph, String email, Set<String> visited, List<String> list) {
list.add(email);
for (String l: graph.get(email)) {
if (visited.add(l)) dfs(graph, l, visited, list);
}
}
}