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>
#include <string.h>

#define DEF_ELEM_MAX 1000001

typedef long long llong;

int A[DEF_ELEM_MAX];
int N;

/////////////////////////////////////////////////////////////////////////////////
//
/////////////////////////////////////////////////////////////////////////////////
void swap(int *x, int *y)
{
	int tmp;
	tmp = *x;
	*x = *y;
	*y = tmp;
	return;
}
/////////////////////////////////////////////////////////////////////////////////
//
/////////////////////////////////////////////////////////////////////////////////
int partition(int *A, int p, int r)
{
	int j;
	int x = A[r];
	int i = p-1;

	for ( j = p ; j < r ; j++ ) {
		if ( A[j] <= x ) {
			i++;
			swap(&A[i], &A[j]);
		}
	}
	swap(&A[i+1], &A[r]);
	return i+1;
}

/////////////////////////////////////////////////////////////////////////////////
//
/////////////////////////////////////////////////////////////////////////////////
int main(void)
{
	int i;
	int k;
//
	scanf("%d", &N);
	for ( i = 0 ; i < N ; i++ ) {
		scanf("%d", &A[i] );
	}

	k = partition(A, 0, N-1);

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

	return 0;
}