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>

void swap(int *x, int *y) {
  int tmp = *x;
  *x = *y;
  *y = tmp;
}

int partition(int a[], int p, int r) {
  int i = p - 1;
  for (int j = p; j <= r - 1; j++) {
    if (a[j] <= a[r]) {
      i++;
      swap(&a[i], &a[j]);
    }
  }
  swap(&a[i+1], &a[r]);
  return i+1;
}
int main() {
  int n;
  scanf("%d", &n);
  int a[n];
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  int v = partition(a, 0, n - 1);
  for (int k = 0; k <= n-1; k++) {
    if (k == v) {
      printf("[%d]", a[k]);
    } else {
      printf("%d", a[k]);
    }
    if (k == n-1) {
      printf("\n");
    } else {
      printf(" ");
    }
  }

  return 0;
}