Rearrange array such that even positioned are greater than odd
Last Updated :
29 Feb, 2024
Given an array arr[] of N elements, sort the array according to the following relations:
- arr[i] >= arr[i – 1], if i is even, ∀ 1 <= i < N
- arr[i] <= arr[i – 1], if i is odd, ∀ 1 <= i < N
Print the resultant array.
Examples:
Input: N = 4, arr[] = {1, 2, 2, 1}
Output: 2 1 2 1
Explanation:
- For i = 1, arr[1] <= arr[0]. So, 1 <= 2.
- For i = 2, arr[2] >= arr[1]. So, 2 >= 1.
- For i = 3, arr[3] <= arr[2]. So, 1 <= 2.
Input: arr[] = {1, 3, 2}
Output: 3 1 2
Explanation:
- For i = 1, arr[1] <= arr[0]. So, 1 <= 3.
- For i = 2, arr[2] >= arr[1]. So, 2 >= 1.
Approach: To solve the problem, follow the below idea:
Observe that array consists of [n/2] even positioned elements. If we assign the largest [n/2] elements to the even positions and the rest of the elements to the odd positions, our problem is solved. Because element at the odd position will always be less than the element at the even position as it is the maximum element and vice versa. Sort the array and assign the first [n/2] elements at even positions.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void assign( int arr[], int N)
{
sort(arr, arr + N);
int ans[N];
int ptr1 = 0, ptr2 = N - 1;
for ( int i = 0; i < N; i++) {
if (i % 2 == 0)
ans[i] = arr[ptr2--];
else
ans[i] = arr[ptr1++];
}
for ( int i = 0; i < N; i++)
cout << ans[i] << " " ;
}
int main()
{
int arr[] = { 1, 2, 2, 1 };
int N = sizeof (arr) / sizeof (arr[0]);
assign(arr, N);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
int cmpfunc( const void * a, const void * b)
{
return (*( int *)a - *( int *)b);
}
void assign( int arr[], int N)
{
qsort (arr, N, sizeof ( int ), cmpfunc);
int ans[N];
int ptr1 = 0, ptr2 = N - 1;
for ( int i = 0; i < N; i++) {
if (i % 2 == 0)
ans[i] = arr[ptr2--];
else
ans[i] = arr[ptr1++];
}
for ( int i = 0; i < N; i++)
printf ( "%d " , ans[i]);
}
int main()
{
int arr[] = { 1, 2, 2, 1 };
int N = sizeof (arr) / sizeof (arr[0]);
assign(arr, N);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
static void assign( int arr[], int N)
{
Arrays.sort(arr);
int ans[] = new int [N];
int ptr1 = 0 , ptr2 = N - 1 ;
for ( int i = 0 ; i < N; i++) {
if (i % 2 == 0 )
ans[i] = arr[ptr2--];
else
ans[i] = arr[ptr1++];
}
for ( int i = 0 ; i < N; i++)
System.out.print(ans[i] + " " );
}
public static void main(String args[])
{
int arr[] = { 1 , 2 , 2 , 1 };
int N = arr.length;
assign(arr, N);
}
}
|
C#
using System;
class GFG {
static void assign( int [] arr, int N)
{
Array.Sort(arr);
int [] ans = new int [N];
int ptr1 = 0, ptr2 = N - 1;
for ( int i = 0; i < N; i++) {
if (i % 2 == 0)
ans[i] = arr[ptr2--];
else
ans[i] = arr[ptr1++];
}
for ( int i = 0; i < N; i++)
Console.Write(ans[i] + " " );
}
public static void Main()
{
int [] arr = { 1, 2, 2, 1 };
int N = arr.Length;
assign(arr, N);
}
}
|
Javascript
<script>
function assign(arr, N)
{
arr.sort();
let ans = [];
let ptr1 = 0, ptr2 = N - 1;
for (let i = 0; i < N; i++) {
if (i % 2 == 0)
ans[i] = arr[ptr2--];
else
ans[i] = arr[ptr1++];
}
for (let i = 0; i < N; i++)
document.write(ans[i] + " " );
}
let arr = [ 1, 2, 2, 1 ];
let N = arr.length;
assign(arr, N);
</script>
|
PHP
<?php
function assign( $arr , $N )
{
sort( $arr );
$ptr1 = 0; $ptr2 = $N - 1;
for ( $i = 0; $i < $N ; $i ++)
{
if ( $i % 2 == 0)
$ans [ $i ] = $arr [ $ptr2 --];
else
$ans [ $i ] = $arr [ $ptr1 ++];
}
for ( $i = 0; $i < $N ; $i ++)
echo ( $ans [ $i ] . " " );
}
$arr = array ( 1, 2, 2, 1 );
$N = sizeof( $arr );
assign( $arr , $N );
?>
|
Python3
def assign(arr, N):
arr.sort()
ans = [ 0 ] * N
ptr1 = 0
ptr2 = N - 1
for i in range (N):
if i % 2 = = 0 :
ans[i] = arr[ptr2]
ptr2 = ptr2 - 1
else :
ans[i] = arr[ptr1]
ptr1 = ptr1 + 1
for i in range (N):
print (ans[i], end = " " )
arr = [ 1 , 2 , 2 , 1 ]
N = len (arr)
assign(arr, N)
|
Time Complexity: O(N * log N), where N is the size of input array arr[].
Auxiliary Space: O(N)
Approach 2: Rearranging array by swapping elements
One other approach is to traverse the array from the first element till N – 1 and swap the element with the next one if the condition is not satisfied. This is implemented as follows:
C++
#include <iostream>
using namespace std;
void rearrange( int arr[], int N)
{
for ( int i = 0; i < N; i += 2) {
if (i > 0 && arr[i - 1] > arr[i])
swap(arr[i - 1], arr[i]);
if (i < N - 1 && arr[i + 1] > arr[i])
swap(arr[i + 1], arr[i]);
}
}
int main()
{
int N = 4;
int arr[] = { 1, 2, 2, 1 };
rearrange(arr, N);
for ( int i = 0; i < N; i++)
cout << arr[i] << " " ;
cout << "\n" ;
return 0;
}
|
Java
import java.io.*;
class GFG {
static void rearrange( int [] arr, int N) {
for ( int i = 0 ; i < N; i += 2 ) {
if (i > 0 && arr[i - 1 ] > arr[i]) {
int temp = arr[i - 1 ];
arr[i - 1 ] = arr[i];
arr[i] = temp;
}
if (i < N - 1 && arr[i + 1 ] > arr[i]) {
int temp = arr[i + 1 ];
arr[i + 1 ] = arr[i];
arr[i] = temp;
}
}
}
public static void main(String[] args) {
int N = 4 ;
int [] arr = { 1 , 2 , 2 , 1 };
rearrange(arr, N);
for ( int i = 0 ; i < N; i++)
System.out.print(arr[i] + " " );
System.out.println();
}
}
|
C#
using System;
class GFG {
static void Rearrange( int [] arr, int N)
{
for ( int i = 0; i < N; i += 2) {
if (i > 0 && arr[i - 1] > arr[i]) {
int temp = arr[i - 1];
arr[i - 1] = arr[i];
arr[i] = temp;
}
if (i < N - 1 && arr[i + 1] > arr[i]) {
int temp = arr[i + 1];
arr[i + 1] = arr[i];
arr[i] = temp;
}
}
}
static void Main()
{
int N = 4;
int [] arr = { 1, 2, 2, 1 };
Rearrange(arr, N);
for ( int i = 0; i < N; i++)
Console.Write(arr[i] + " " );
Console.WriteLine();
}
}
|
Javascript
function rearrange(arr, N) {
for (let i = 0; i < N; i += 2) {
if (i > 0 && arr[i - 1] > arr[i]) {
[arr[i - 1], arr[i]] = [arr[i], arr[i - 1]];
}
if (i < N - 1 && arr[i + 1] > arr[i]) {
[arr[i], arr[i + 1]] = [arr[i + 1], arr[i]];
}
}
}
const N = 4;
const arr = [1, 2, 2, 1];
rearrange(arr, N);
console.log(arr.join( " " ));
|
Python3
def rearrange(arr, N):
for i in range ( 0 , N, 2 ):
if i > 0 and arr[i - 1 ] > arr[i]:
arr[i - 1 ], arr[i] = arr[i], arr[i - 1 ]
if i < N - 1 and arr[i + 1 ] > arr[i]:
arr[i], arr[i + 1 ] = arr[i + 1 ], arr[i]
N = 4
arr = [ 1 , 2 , 2 , 1 ]
rearrange(arr, N)
for i in range (N):
print (arr[i], end = " " )
print ()
|
Time Complexity: O(N), where N is the size of input array arr[].
Auxiliary Space: O(1)
Share your thoughts in the comments
Please Login to comment...