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>
#define MAX 100000
 
int A[MAX];
int partition();
 
 
int main(){

  int i;
  int n,q;
 
  scanf("%d",&n);

  for( i=0 ; i < n ; i++ ){

    scanf("%d",&A[i]);

  }

  q = partition( 0 , n-1 );
 
  for( i = 0; i < n; i++ ){

    if(i == q) printf("[");
 
      printf( "%d", A[i]);
 
      if(i == q) printf("]");
 
      if(i != n-1) printf(" ");
  }
  printf("\n");
 
  return 0;
}
 
int partition(int p,int r)
{
  int i,j;
  int x;
  int s;
 
  x = A[r];
  i = p-1;

  for(j = p; j < r; j++){

    if(A[j]<= x){

      i++;
      s = A[i];
      A[i] = A[j];
      A[j] = s;
    }
  }
      s = A[i+1];
      A[i+1] = A[r];
      A[r] = s;
  
    return i+1;
  }