Problem:
Partition

Quick sort is based on the Divide-and-conquer approach. In QuickSort(A, p, r), first, a procedure Partition(A, p, r) divides an array A[p..r] into two subarrays A[p..q-1] and A[q+1..r] such that each element of A[p..q-1] is less than or equal to A[q], which is, inturn, less than or equal to each element of A[q+1..r]. It also computes the index q.

In the conquer processes, the two subarrays A[p..q-1] and A[q+1..r] are sorted by recursive calls of QuickSort(A, p, q-1) and QuickSort(A, q+1, r).

Your task is to read a sequence A and perform the Partition based on the following pseudocode:

Partition(A, p, r)
1 x = A[r]
2 i = p-1
3 for j = p to r-1
4 do if A[j] <= x
5 then i = i+1
6 exchange A[i] and A[j]
7 exchange A[i+1] and A[r]
8 return i+1

Note that, in this algorithm, Partition always selects an element A[r] as a pivot element around which to partition the array A[p..r].

Input

The first line of the input includes an integer n, the number of elements in the sequence A.

In the second line, Ai (i = 1,2,...,n), elements of the sequence are given separated by space characters.

Output

Print the sorted sequence. Two contiguous elements of the sequence should be separated by a space character. The element which is selected as the pivot of the partition should be indicated by [  ].

Constraints

1 ≤ n ≤ 100,000

0 ≤ Ai ≤ 100,000

Sample Input 1

12
13 19 9 5 12 8 7 4 21 2 6 11

Sample Output 1

9 5 8 7 4 2 6 [11] 21 13 19 12

Solution in C (judge verdict: Accepted)

#include<stdio.h>

int main(){
  //counter
  int i,j;

  //input
  int n;
  scanf("%100000d",&n);

  int A[n];
  for(i=0;i<n;i++){
    scanf("%100000d",&A[i]);
  }


  
  //PartitionSort

  int p = 0;
  int r = n-1;
  int x = A[r];
  int temp;

  i = p-1;

  for(j=p;j<=(r-1);j++){
    if(A[j]<=x){
      i++;
      //exchange A[i] and A[j]
      temp = A[i]; A[i] = A[j]; A[j] = temp;
    }
  }
  i++;
  //exchange A[i] and A[r]
  temp = A[r]; A[r] = A[i]; A[i] = temp;
  
      
  int k = i;

  //Output

  for(i=0;i<n-1;i++){
    if(i == k){
      printf("[%d] ",x);
    }else{
      printf("%d ",A[i]);
  }
  }
    if(n-1 == k){
      printf("[%d]\n",A[n-1]);
    }
       else{
  printf("%d\n",A[n-1]);
       }

  
  return 0;
}