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AccountsMerge.cpp
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AccountsMerge.cpp
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// Source : https://leetcode.com/problems/accounts-merge/
// Author : Hao Chen
// Date : 2019-03-29
/*****************************************************************************************************
*
* 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:
* "[email protected]"], ["Mary", "[email protected]"]]
*
* Explanation:
* The first and third John's are the same person as they have the common email "[email protected]".
* 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', '[email protected]'],
* ['John', '[email protected]'],
*
* 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].
******************************************************************************************************/
//Bad Performance Solution
class Solution_Time_Limit_Exceeded {
public:
// We can orginze all relevant emails to a chain,
// then we can use Union Find algorithm
// Besides, we also need to map the relationship between name and email.
vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
unordered_map<string, string> emails_chains; // email chains
unordered_map<string, string> names; // names to email chains' head
//initialization
for(int i = 0 ; i<accounts.size();i++) {
auto& account = accounts[i];
auto& name = account[0];
for (int j=1; j<account.size(); j++) {
auto& email = account[j];
if ( names.find(email) == names.end() ) {
emails_chains[email] = email;
names[email] = name;
}
join(emails_chains, account[1], email);
}
}
//reform the emails
unordered_map<string, set<string>> res;
for( auto& acc : accounts ) {
string e = find(emails_chains, acc[1]);
res[e].insert(acc.begin()+1, acc.end());
}
//output the result
vector<vector<string>> result;
for (auto pair : res) {
vector<string> emails(pair.second.begin(), pair.second.end());
emails.insert(emails.begin(), names[pair.first]);
result.push_back(emails);
}
return result;
}
string find(unordered_map<string, string>& emails_chains,
string email) {
while( email != emails_chains[email] ){
email = emails_chains[email];
}
return email;
}
bool join(unordered_map<string, string>& emails_chains,
string& email1, string& email2) {
string e1 = find(emails_chains, email1);
string e2 = find(emails_chains, email2);
if ( e1 != e2 ) emails_chains[e1] = email2;
return e1 == e2;
}
};
//
// Performance Tunning
// -----------------
//
// The above algorithm need to do string comparison, it causes lots of efforts
// So, we allocated the ID for each email, and compare the ID would save the time.
//
// Furthermore, we can use the Group-Email-ID instead of email ID,
// this would save more time.
//
class Solution {
public:
// We can orginze all relevant emails to a chain,
// then we can use Union Find algorithm
// Besides, we also need to map the relationship between name and email.
vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {
unordered_map<string, int> emails_id; //using email ID for union find
unordered_map<int, int> emails_chains; // email chains
unordered_map<int, string> names; // email id & name
//initialization & join
for(int i = 0 ; i<accounts.size();i++) {
// using the account index as the emails group ID,
// this could simplify the emails chain.
emails_chains[i] = i;
auto& account = accounts[i];
auto& name = account[0];
for (int j=1; j<account.size(); j++) {
auto& email = account[j];
if ( emails_id.find(email) == emails_id.end() ) {
emails_id[email] = i;
names[i] = name;
}else {
join( emails_chains, i, emails_id[email] );
}
}
}
//reform the emails
unordered_map<int, set<string>> res;
for(int i=0; i<accounts.size(); i++) {
int idx = find(emails_chains, i);
res[idx].insert(accounts[i].begin()+1, accounts[i].end());
}
//output the result
vector<vector<string>> result;
for (auto pair : res) {
vector<string> emails( pair.second.begin(), pair.second.end() );
emails.insert(emails.begin(), names[pair.first]);
result.push_back(emails);
}
return result;
}
int find(unordered_map<int, int>& emails_chains, int id) {
while( id != emails_chains[id] ){
id = emails_chains[id];
}
return id;
}
bool join(unordered_map<int, int>& emails_chains, int id1, int id2) {
int e1 = find(emails_chains, id1);
int e2 = find(emails_chains, id2);
if ( e1 != e2 ) emails_chains[e1] = e2;
return e1 == e2;
}
};