Problem:
Score : 300 points

Problem Statement

Given is a sequence of integers A_1, A_2, ..., A_N.
If its elements are pairwise distinct, print YES; otherwise, print NO.

Constraints

2 ≤ N ≤ 200000

1 ≤ A_i ≤ 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

If the elements of the sequence are pairwise distinct, print YES; otherwise, print NO.

Sample Input 1

5
2 6 1 4 5

Sample Output 1

YES

The elements are pairwise distinct.

Sample Input 2

6
4 1 3 1 6 2

Sample Output 2

NO

The second and fourth elements are identical.

Sample Input 3

2
10000000 10000000

Sample Output 3

NO

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;
import static java.lang.System.out;

public class Main{
    static MyReader in = new MyReader();

    public static void main(String[] args){
        int N = in.i();
        int[] A = in.ii();
        HashSet<Integer> hs = new HashSet<>();
        for(int i = 0; i < N; i++){
            hs.add(A[i]);
        }        
        out.println(hs.size() == N ? "YES" : "NO");
    }

    static class MyReader extends BufferedReader{
        MyReader(){
            super(new InputStreamReader(System.in));
        }

        String s(){
            try{return readLine();}catch(IOException e){return "";}
        }

        String[] ss(){
            return s().split(" ");
        }

        int i(){
            return Integer.parseInt(s());
        }

        int[] ii(){
            String[] ss = ss();
            int[] ii = new int[ss.length];
            for(int j = 0; j < ss.length; j++) ii[j] = Integer.parseInt(ss[j]);
            return ii;
        }

        ArrayList<Integer> li(){
            String[] ss = ss();
            ArrayList<Integer> li = new ArrayList<>();
            for(int j = 0; j < ss.length; j++) li.add(Integer.parseInt(ss[j]));
            return li;
        }

        long l(){
            return Long.parseLong(s());
        }

        long[] ll(){
            String[] ss = ss();
            long[] ll = new long[ss.length];
            for(int j = 0; j < ss.length; j++) ll[j] = Long.parseLong(ss[j]);
            return ll;
        }
    }
}