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()
{
    int i, j, n, x, p, q, r, s, count = 0;
    scanf("%d", &n);
    int A[n];
    for(i = 0; i < n; i++) scanf("%d", &A[i]);
    
    p = 0;
    r = n - 1;
    x = A[r];
    i = p-1;
    for (j = p; j <= r; j++){
     if (A[j] <= x){
       i = i + 1;
        s = A[i];
       A[i] = A[j];
       A[j] = s;
   //s = A[i+1];
   //A[i+1] = A[r];
   //A[r] = s;
     }
    }
    
    for(i = 0; i < n; i++){
        if(A[i] == x) count++;
    }
    
    for(i = 0; i < n - 1; i++){
        if(A[i] == x) count--;
    if(count == 0){
        if(A[i] == x) {
            printf("[%d] ", A[i]);
            continue;
    }
    }
    printf("%d ", A[i]);
    }
    printf("%d\n", A[n-1]);
    return 0;
}