Given an array of size n, arrange the first k elements of the array in ascending order and the remaining n-k elements in descending order.
Examples:
Input: arr[] = {5, 4, 6, 2, 1, 3, 8, 9, -1}, k = 4
Output: 2 4 5 6 9 8 3 1 -1
Input: arr[] = {5, 4, 6}, k = 2
Output: 4 5 6
Algorithm:
- Store the first k elements in an array and sort that in ascending order.
- Store the remaining n-k elements in an array and sort that in descending order.
- Merge the two arrays by adding the elements from the second array in reverse order.
C++
#include <bits/stdc++.h>
using namespace std;
void printOrder( int arr[], int n, int k)
{
int len1 = k, len2 = n - k;
int arr1[k], arr2[n - k];
for ( int i = 0; i < k; i++)
arr1[i] = arr[i];
for ( int i = k; i < n; i++)
arr2[i - k] = arr[i];
sort(arr1, arr1 + len1);
sort(arr2, arr2 + len2);
for ( int i = 0; i < n; i++) {
if (i < k)
arr[i] = arr1[i];
else {
arr[i] = arr2[len2 - 1];
len2--;
}
}
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
}
int main()
{
int arr[] = { 5, 4, 6, 2, 1, 3, 8, 9, -1 };
int k = 4;
int n = sizeof (arr) / sizeof (arr[0]);
printOrder(arr, n, k);
return 0;
}
|
Java
import java.util.*;
class GFG {
static void printOrder( int [] arr, int n, int k)
{
int len1 = k, len2 = n - k;
int [] arr1 = new int [k];
int [] arr2 = new int [n - k];
for ( int i = 0 ; i < k; i++)
arr1[i] = arr[i];
for ( int i = k; i < n; i++)
arr2[i - k] = arr[i];
Arrays.sort(arr1, 0 , k);
Arrays.sort(arr2, 0 , n - k);
for ( int i = 0 ; i < n; i++) {
if (i < k)
arr[i] = arr1[i];
else {
arr[i] = arr2[len2 - 1 ];
len2--;
}
}
for ( int i = 0 ; i < n; i++) {
System.out.print(arr[i] + " " );
}
}
public static void main(String[] args)
{
int arr[] = { 5 , 4 , 6 , 2 , 1 , 3 , 8 , 9 , - 1 };
int k = 4 ;
int n = arr.length;
printOrder(arr, n, k);
}
}
|
Python3
def printOrder(arr, n, k):
len1 = k
len2 = n - k
arr1 = [ 0 ] * k
arr2 = [ 0 ] * (n - k)
for i in range (k):
arr1[i] = arr[i]
for i in range (k, n):
arr2[i - k] = arr[i]
arr1.sort()
arr2.sort()
for i in range (n):
if (i < k):
arr[i] = arr1[i]
else :
arr[i] = arr2[len2 - 1 ]
len2 - = 1
for i in range (n):
print (arr[i], end = " " )
if __name__ = = "__main__" :
arr = [ 5 , 4 , 6 , 2 , 1 ,
3 , 8 , 9 , - 1 ]
k = 4
n = len (arr)
printOrder(arr, n, k)
|
C#
using System;
class GFG {
static void printOrder( int [] arr,
int n, int k)
{
int len2 = n - k;
int [] arr1 = new int [k];
int [] arr2 = new int [n - k];
for ( int i = 0; i < k; i++)
arr1[i] = arr[i];
for ( int i = k; i < n; i++)
arr2[i - k] = arr[i];
Array.Sort(arr1, 0, k);
Array.Sort(arr2, 0, n - k);
for ( int i = 0; i < n; i++) {
if (i < k)
arr[i] = arr1[i];
else {
arr[i] = arr2[len2 - 1];
len2--;
}
}
for ( int i = 0; i < n; i++) {
Console.Write(arr[i] + " " );
}
}
public static void Main()
{
int [] arr = { 5, 4, 6, 2, 1,
3, 8, 9, -1 };
int k = 4;
int n = arr.Length;
printOrder(arr, n, k);
}
}
|
PHP
<?php
function printOrder( $arr , $n , $k )
{
$len1 = $k ;
$len2 = $n - $k ;
$arr1 = array_fill (0, $k , 0);
$arr2 = array_fill (0, ( $n - $k ), 0);
for ( $i = 0; $i < $k ; $i ++)
$arr1 [ $i ] = $arr [ $i ];
for ( $i = $k ; $i < $n ; $i ++)
$arr2 [ $i - $k ] = $arr [ $i ];
sort( $arr1 );
sort( $arr2 );
for ( $i = 0; $i < $n ; $i ++)
if ( $i < $k )
$arr [ $i ] = $arr1 [ $i ];
else
{
$arr [ $i ] = $arr2 [ $len2 - 1];
$len2 -= 1;
}
for ( $i = 0; $i < $n ; $i ++)
print ( $arr [ $i ] . " " );
}
$arr = array ( 5, 4, 6, 2, 1, 3, 8, 9, -1 );
$k = 4;
$n = count ( $arr );
printOrder( $arr , $n , $k );
?>
|
Javascript
<script>
function printOrder(arr, n, k)
{
let len1 = k, len2 = n - k;
let arr1 = new Array(k);
let arr2 = new Array(n - k);
for (let i = 0; i < k; i++)
arr1[i] = arr[i];
for (let i = k; i < n; i++)
arr2[i - k] = arr[i];
arr1.sort( function (a, b){ return a - b;});
arr2.sort( function (a, b){ return a - b;});
for (let i = 0; i < n; i++)
{
if (i < k)
arr[i] = arr1[i];
else
{
arr[i] = arr2[len2 - 1];
len2--;
}
}
for (let i = 0; i < n; i++)
{
document.write(arr[i] + " " );
}
}
let arr = [ 5, 4, 6, 2, 1, 3, 8, 9, -1 ];
let k = 4;
let n = arr.length;
printOrder(arr, n, k);
</script>
|
Output:
2 4 5 6 9 8 3 1 -1
Time Complexity: O(N * log(N))
Auxiliary Space: O(N)
Efficient Approach: The idea is simple, sort the first k elements in increasing order and remaining n-k elements in decreasing using library function.
C++
#include <bits/stdc++.h>
using namespace std;
void printOrder( int arr[], int n, int k)
{
sort(arr, arr + k);
sort(arr + k, arr + n, greater< int >());
}
int main()
{
int arr[] = { 5, 4, 6, 2, 1, 3, 8, 9, -1 };
int k = 4;
int n = sizeof (arr) / sizeof (arr[0]);
printOrder(arr, n, k);
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
return 0;
}
|
Java
import java.util.*;
public class SortExample {
static void printOrder(Integer[] arr, int k)
{
int n = arr.length;
Arrays.sort(arr, 0 , k);
Arrays.sort(arr, k, n, Collections.reverseOrder());
}
public static void main(String[] args)
{
Integer[] arr = { 5 , 4 , 6 , 2 , 1 , 3 , 8 , 9 , - 1 };
int k = 4 ;
printOrder(arr, k);
System.out.printf( "%s" , Arrays.toString(arr));
}
}
|
Python3
def printOrder(arr, n, k):
a = arr[ 0 :k];
a.sort();
b = arr[k:n];
b.sort();
b.reverse();
return a + b;
arr = [ 5 , 4 , 6 , 2 , 1 , 3 , 8 , 9 , - 1 ];
k = 4 ;
n = len (arr);
arr = printOrder(arr, n, k);
for i in range (n):
print (arr[i], end = " " );
|
C#
using System;
public class SortExample {
static void printOrder( int [] arr, int k)
{
int n = arr.Length;
Array.Sort(arr, 0, k);
Array.Sort(arr, k, n - k);
Array.Reverse(arr, k, n - k);
}
public static void Main(String[] args)
{
int [] arr = { 5, 4, 6, 2, 1, 3, 8, 9, -1 };
int k = 4;
printOrder(arr, k);
Console.Write( "{0}" , String.Join( " " , arr));
}
}
|
PHP
<?php
function printOrder( $arr , $n , $k )
{
$a = array_slice ( $arr , 0, $k );
sort( $a );
$b = array_slice ( $arr , $k , $n );
sort( $b );
$b = array_reverse ( $b );
unset( $arr );
$arr = $a ;
return array_merge ( $arr , $b );
}
$arr = array ( 5, 4, 6, 2, 1, 3, 8, 9, -1 );
$k = 4;
$n = count ( $arr );
$arr =printOrder( $arr , $n , $k );
for ( $i = 0; $i < $n ; $i ++)
echo $arr [ $i ]. " " ;
?>
|
Javascript
<script>
function printOrder(arr, k)
{
let n = arr.length;
arr = arr.slice(0, k).sort(
function (a, b){ return a - b;}).concat(
arr.slice(k, n).sort( function (a, b){ return b - a;}));
return arr;
}
let arr = [ 5, 4, 6, 2, 1, 3, 8, 9, -1 ];
let k = 4;
arr = printOrder(arr, k);
document.write(arr.join( " " ));
</script>
|
Output:
2 4 5 6 9 8 3 1 -1
Time Complexity: O(N * log(N))
Auxiliary Space: O(1)
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Last Updated :
23 Jun, 2022
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