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 C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>

void q_sort(long numbers[], int left, int right);

int main(void){
  int j,N,K,i,min;
  long A[200000]={};
  long ans;
  if(scanf("%d",&N)>1){
    exit(0);
  }
  for(i=0;i<N;i++){
   if(scanf("%ld",&A[i])>1){
    exit(0);
   }
  }
  q_sort(A,0,N-1);
  for(i=0;i<N;i++){
      if(A[i]==A[i+1]){
	printf("NO");
	goto out;
      }
      if(i==N-1){
      printf("YES");
    }
  }
 out:
  return 0;
}

void q_sort(long numbers[], int left, int right)
{
    int l_hold, r_hold;
    long pivot,pa;

    l_hold = left;
    r_hold = right;
    pivot = numbers[left];
    while (left < right)
    {
        while ((numbers[right] >= pivot) && (left < right))
            right--;
        if (left != right)
        {
            numbers[left] = numbers[right];
            left++;
        }
        while ((numbers[left] <= pivot) && (left < right))
            left++;
        if (left != right)
        {
            numbers[right] = numbers[left];
            right--;
        }
   }
    numbers[left] = pivot;
    pivot = left;
    left = l_hold;
    right = r_hold;
    if (left < pivot)
      q_sort(numbers, left, pivot-1);
    if (right > pivot)
      q_sort(numbers, pivot+1, right);
}