Program to sort the array based on pivot value (Quick sort)

 Partition the given array based on pivot value

Given a pivot x and list 1st,partition the list into three parts.The first part contains

all the elements in the 1st that are less than x.The second part contains all elements 

in the list that are equal to x.The third part contains all elements in the lst

that are larger than x.Ordering within a part can be arbitary.The algorithm result in

minimum swaps.

I/P:

7

9 12 3 5 14 10 10

10

O/P:

3 5 9 10 10 12 14

PROGRAM:

package Arrays;

import java.util.*;

public class pivotvalue {

    public static void main(String args[])

    {

        Scanner sc=new Scanner(System.in);

        int n=sc.nextInt();

        int arr[]=new int[n];

        for(int i=0;i<n;i++)

        {

            arr[i]=sc.nextInt();

        }

        int pivot=sc.nextInt();

        int a1[]=new int[n];

        int a2[]=new int[n];

        int a3[]=new int[n];

        int first=0,second=0,third=0;

        for(int i=0;i<n;i++)

        {

            if(arr[i]<pivot)

            {

               a1[first]=arr[i];   

               first++;

            }

            else if(arr[i]==pivot)

            {

                a2[second]=arr[i];

                second++;

            }

            else if(arr[i]>pivot){

                a3[third]=arr[i];

                third++;

            }

        }

        Arrays.sort(a1,0,first);

        Arrays.sort(a3,0,third);

        for(int i=0;i<first;i++)

        {

            System.out.print(a1[i]+" ");

        }

        for(int i=0;i<second;i++)

        {

            System.out.print(a2[i]+" ");

        }

        for(int i=0;i<third;i++)

        {

            System.out.print(a3[i]+" ");

        }

        

    }

    

}