I wonder why Arrays.sort wont work on strings..

You should use "[a-z]+@gmail\\.com$"; instead. Because "." in regex means any character except newline

Hello,can you tell me how it avoids duplicate emails.

Yea I hate libriaties too, I always try to make my own algorithms to understand how things might work. Even if I use libriary, I try to look at it's sourse code, and read about it much to understand how its working.

teminate the scanner at the end using scanner.close();

The "import java.util.*;" statement should have imported the Collections and regex libraries. I don't understand why we are requires to write separate statements for importing the Collections and Regex libraries.

Just a little comment on the Regex for anyone interested. The constraint given about the email addresses says "Each of the email IDs consists of lower case letters [a-z], @ and . only". Based on this definition, a better regex is "^[a-z\.]+@gmail\.com$".

Explanation: The ^ at the start is an anchor (so nothing is allowed to come before it), then [a-z\.] includes the lower case letters and . and the + means anything in that group 1 or more times.

Edit: You'll need to use two slashes where it's showing one.

Suggestion: Use std::multiset instead of std::vector. Then you don't need to do sort, like this(just gonna re-write your code a bit differently):

#include <set>
#include <iostream>
using namespace std;

int main()
{
    int N;
    cin >> N;
    regex e(".+@gmail\\.com$");
    multiset<string> db;
    for(int a0 = 0; a0 < N; a0++)
    {
        string firstName;
        string emailID;
        cin >> firstName >> emailID;
        if(regex_match(emailID,e))
        {
            db.insert(firstName);
        }
    }
    // just a fancy and less robust way of iterating in 
    // C++11 (and later)
    for(auto& it : db) 
    { 
        cout << it << endl; 
    }
    return 0;
}

Regexp in c++ working? In gcc 5.3.1 wasnt..

Also:

vector<string> names;
for(int a0 = 0; a0 < N; a0++){
    string firstName;
    string emailID;
    cin >> firstName >> emailID;
    size_t t = emailID.find("@");
    if (t != string::npos && emailID.substr(t) == "@gmail.com" && emailID.substr(0,t).find(firstName) != string::npos)
        names.push_back(firstName);
}
sort(names.begin(), names.end());
for(vector<string>::iterator i = names.begin(); i != names.end(); ++i)
    cout << *i << endl;

why do you need the "\\" in

regex e(".+@gmail\\.com$")

Can you please explain me this code?

Can anyone tell me why this regex exp doesn't work

regex r("[a-z]+@gmail\\.com$");

The constraints does say Each of the email IDs consists of lower case letters [a-z] , @ and . only.

Wow, great way

if __ in __:

	for the win

:) :) :)

I agree. I am terrible at re and yet I finished in five minutes typing on an iPhone. I have spend decades programming around regex, and recently spent many hours getting a handle on lookahead and lookbehind. Both disappointed and relieved not to spend my hackos on weird edge cases to test my patterns.

true, also who the hell needs a regex if you have

str.endswith("@gmail.com")

as they end all the same way :D

This is a classic task where I end up with a stupid one-liner like:

print("\n".join(sorted([l.split()[0] for l in [input() for _ in range(int(input()))] if '@gmail.com' in l])))

You could get even more Pythonic (also conforming to PEP 8)

import re

lst=[]
for a0 in range(int(input())):
    firstName, emailID = [str(s) for s in input().split()]
    if re.search('@gmail\.com$', emailID):
   	    lst.append(firstName)
print(*sorted(lst), sep='\n')

Suggest moving the rgx instantiation to outside the for() loop, instantiating once instead of for each entry. ie:

static void Main(String[] args)
  {
    int N = Convert.ToInt32(Console.ReadLine());
    Regex rgx = new Regex(pattern);
    
    for(int a0 = 0; a0 < N; a0++)
      ...

I like your use of printing *names with a newline separator.